Download Sequences, Series, and Mathematical Induction

Sequences, Series, and
Mathematical Induction
Raja Almukahhal
Larame Spence
Mara Landers
Nick Fiori
Art Fortgang
Melissa Vigil
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Printed: October 2, 2013
AUTHORS
Raja Almukahhal
Larame Spence
Mara Landers
Nick Fiori
Art Fortgang
Melissa Vigil
www.ck12.org
Chapter 1. Sequences, Series, and Mathematical Induction
C HAPTER
1
Sequences, Series, and
Mathematical Induction
C HAPTER O UTLINE
1.1
Recursive Formulas
1.2
Explicit Formulas
1.3
Sum Notation and Properties of Sigma
1.4
Series Sums and Gauss’ Formula
1.5
Problem Solving with Series Sums
1.6
Inductive Proofs
1.7
Induction and Factors
1.8
Induction and Inequalities
1.9
Sums of Finite Geometric Series
1.10
Sums of Infinite Geometric Series
1.11
Factorials and Combinations
1.12
Binomial Theorem and Expansions
Introduction
Sequences and series are really just patterns of numbers, and numerical patterns are everywhere. Patterns are found
in the growth of trees, the enrollment in an online group, the movement of a bird, the layout of atoms in a crystal,
and a million other places.
Understanding how numerical patterns work is a fundamental step toward predicting future events. Imagine using
numbers to predict the best way to throw a baseball, or pick the best pitcher out of a collection of free agents.
Number series appear in the way people choose to buy products, or donate money.
In this chapter, you will explore the topics above and will learn to both use formulas designed to help you create
series, and create formulas from series in order to predict later members of sequences.
1
1.1. Recursive Formulas
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1.1 Recursive Formulas
Here you will learn about sequences and series, and will explore the concept of a "recursive formula".
Have you ever had one of those "advent calendars" around the Dec. holidays? Commonly they have a little piece of
chocolate for each day to count down the days until Xmas. If the calendar starts on the 1st, there are 25 candies at
the beginning of the month, then 24 remaining on the 2nd, 23 on the 3rd, and so forth.
The number of pieces remaining after each day is a series: {25, 24, 23...}. How could you identify how many are
left on any specific day of the month? Is there a general formula of calculating the candy remaining on any date?
Watch This
Embedded Video:
MEDIA
Click image to the left for more content.
- PatrickJMT: Recursive Sequences
Guidance
A sequence is an ordered list of objects. The simplest way to represent a sequence is by listing some of its terms.
The sequence of odd, positive integers is shown here:
TABLE 1.1:
1, 3, 5, 7 ...
In this lesson you will learn to represent a sequence recursively, which means that you need to know the previous
term in order to find the next term in the sequence.
Consider the sequence shown above. What is the next term?
As long as you are familiar with the odd integers (i.e., you can count in 2’s) you can figure out that the next term is
9. If we want to describe this sequence in general, we can do so by stating what the first term is, and then by stating
the relationship between successive terms. When we represent a sequence by describing the relationship between its
successive terms, we are representing the sequence recursively.
The terms in a sequence are often denoted with a variable and a subscript. All of the terms in a given sequence are
written with the same variable, and increasing subscripts. So we might list terms in a sequence as a1 , a2 , a3 , a4 , a5
...
We can use this notation to represent the example above. This sequence is defined as follows:
2
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Chapter 1. Sequences, Series, and Mathematical Induction
TABLE 1.2:
a1 = 1
an = an−1 + 2
At first glance this notation may seem confusing. What is important to keep in mind is that the subscript of a term
represents its “place in line.” So an just means the nth term in the sequence. The term an−1 just means the term
before an . In the sequence of odd numbers above, a1 = 1, a2 = 3, a3 = 5, a4 = 7, a5 = 9 and so on. If, for example,
we wanted to find a10 , we would need to find the 9th term in the sequence first. To find the 9th term we need to find
the 8th term, and so on, back to a term that we know.
Example A
For the sequence of odd numbers, list a6 , a7 , a8 , a9 , and a10
Solution Each term is two more than the previous term.
a6 = a5 + 2 = 9 + 2 = 11
a7 = a6 + 2 = 11 + 2 = 13
a8 = a7 + 2 = 13 + 2 = 15
a9 = a8 + 2 = 15 + 2 = 17
a10 = a9 + 2 = 17 + 2 = 19
The sequence of odd numbers is linear, and it is referred to as an arithmetic sequence. Every arithmetic sequence
has a common difference, or a constant difference between each term. (The common difference is analogous to the
slope of a line.) The sequence of odd numbers has a common difference of 2 because for all n, an - an−1 = 2.
Finding terms in this sequence is relatively straightforward, as the pattern is familiar. However, this would clearly
be tedious if you needed to find the 100th term.
Example B
Find the 5th term for the sequence:
TABLE 1.3:
t1 = 3
tn = 2tn−1
Solution
t5 = 48
TABLE 1.4:
t1 = 3
tn = 2 × tn−1
→
t2 = 2t1 = 2 × 3 = 6
t3 = 2t2 = 2 × 6 = 12
t4 = 2t3 = 2 × 12 = 24
t5 = 2t4 = 2 × 24 = 48
This example is a geometric sequence. Every geometric sequence has a common ratio, which is 2 in this example,
tn
because for all n, tn−1
= 2. The terms of a geometric sequence follow an exponential pattern.
3
1.1. Recursive Formulas
www.ck12.org
Example C
Find the 4th term for the sequence:
TABLE 1.5:
b1 = 2
bn = (bn−1 )2 + 1
Solution
b4 = 677
TABLE 1.6:
b1 = 2
b2 = (b1 )2 + 1 = 22 + 1 = 4 + 1 = 5
b3 = (b2 )2 + 1 = 52 + 1 = 25 + 1 =
26
b4 = (b3 )2 + 1 = 262 + 1 = 676 + 1 =
677
→
bn = (bn−1 )2 + 1
This sequence is neither arithmetic nor geometric, though its values follow a cubic pattern. As you can see from just
a few terms here, the terms in a sequence can grow quickly.
For any of these sequences, as noted above, determining more than a few values by hand can be time consuming.
In another lesson, we will introduce explicit formulas, which can be used to define a sequence in a way that makes
finding the nth term faster.
TABLE 1.7:
Concept question wrap-up The remaining candy in an advent calendar is a standard arithmetic sequence, and can
be described as an - an−1 = 1
Vocabulary
To represent a sequence recursively means that you need to know the previous term in order to find the next term in
the sequence.
The sequence of odd numbers is linear because it would be a straight line if graphed on a rectangular graph.
An arithmetic sequence has a common difference, or the same number of values between each term.
A geometric sequence has a common ratio, or uses the same multiple to get from one term to another.
Explicit formulas describe each term directly, rather than defining each term by the one before.
Guided Practice
Questions
1) Let’s look at another recursive formula and discover the next 5 terms:
Given:a1 = 2 and an = 3an−1 + 3.
a2 = 3(2) + 3 = 9
4
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Chapter 1. Sequences, Series, and Mathematical Induction
a3 = 3(9) + 3 = 30
a4 = 3(30) + 3 = 93
a5 = 3(93) + 3 = 282
a6 = 3(282) + 3 = 849
So our answer is: 9, 30, 93, 282 and 849
2) Write the next 5 terms of the sequence. Given: a1 = −4a2 = −4 and an = 2an−1 + an−2
a2 = 2(−4) + (−4) = −12
a3 = 2(−12) + (−4) = −28
a4 = 2(−28) + (−12) = −68
a5 = 2(−68) + (−28) = −164
a6 = 2(−164) + (−68) = −396
So our answer is: -12, -28, -68, -164, and -396.
3) Write a recursive formula that fits the following sequence:
1, 5, 9, 13, 17
In this problem we deduct that each term differs by the same amount. Once, we identify the difference of each term,
+4 each time in this case, then we know that the sequence requires that we add the same amount (4) to each term.
We write that as: an = an−1 + 4
4) Given the following sequence, write a recursive formula, then find the next three numbers in the series.
3, -4, -1, -5, -6, -11, -17
Getting to the next term is not going to be as easy as the previous example. We need to examine the number sequence
more closely to solve this problem.
In this sequence, the Fibonacci Series was applied. What this means is that the two previous terms were added
together to get the next term in the sequence.
It is written as: an = an−1 + an−2
Now that we know the formula, we can find the next three numbers in the sequence:
a8 = (−11) + (−17) = −28
a9 = (−17) + (−28) = −45
a10 = (−28) + (−45) = −73
So the next three numbers in the sequence are: -28, -45, and -73.
Practice
Recursive Definitions:
1.
2.
3.
4.
A sequence in which you know the previous term in order to find the next term is:
Why is the sequence of odd numbers linear?
Which type of sequence has a common difference?
A sequence that uses the same multiple to get from one term to another is:
5
1.1. Recursive Formulas
www.ck12.org
5. Find the value of a6 , given the sequence defined as: a1 = 4 an = 5an−1
6. Find the value of a5 , given the sequence defined as: a1 = 32 an = (1/2)an−1
7. Find the value of an−1 , given the sequence defined as: a1 = 1 an = 3an−1 -n
Using the given recursive formulas, identify the next 5 terms in the sequences that follow:
8.
9.
10.
11.
12.
a1 = −2a2 = 1 and an = 3an−1 − 5an−2
a1 = −2 and an = 3an1
a1 = 3a2 = −2 and an = −5an−1 + an−2
a1 = 1 and an = 4an−1
a1 = −4a2 = 1 and an = −an−1 + an−2
Given the following sequence of numbers find the recursive formula
13. 1, 5, 9, 13, 17
14. -1, 3, 2, 5, 7, 12, 19
15. -4, 16, -64, 256, -1024
Given the following sequence of numbers find the recursive formula and the next three numbers in the sequence.
16.
17.
18.
19.
20.
21.
22.
23.
6
1, -1, 1, -1, 1
-5, -1, -6, -7, -13, -20, -33
1, - 3, 9, -27, 81
-3, -4, -7, -11, -18, -29, -47
-1, -5, -9, -13, -17
Write the next three terms of the sequence: an = (−1)n · 5an−1
Given the formula: an = 4n − 1, is the number 27 a term in the sequence of numbers?
Given the formula: an = 4n − 1 is the number 97 a term in the sequence of numbers?
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Chapter 1. Sequences, Series, and Mathematical Induction
1.2 Explicit Formulas
Here you will learn about arithmetic and geometric sequences, and you will explore using explicit formulas, which
allow you to find a term without knowing the one before it.
Rachel and Elaina have started a website where they debate the best color of hair dye. The show is really popular
and the visitors to their website are increasing very rapidly. They figure that memberships are increasing by about
500 people every three days.
At this rate, how many members will they have on the 48th day? How many days will it be before they reach 25,000
members?
Watch This
TABLE 1.8:
Arithmetic Sequences: Watch this v
ideo on arithmetic sequences at this
link.
Geometric Sequences: Learn about
geometric sequences by watching
the video at this link.
Guidance
When we represent a sequence with a formula that lets us find any term in the sequence without knowing any other
terms, we are representing the sequence explicitly.
Given a recursive definition of an arithmetic or geometric sequence, you can always find an explicit formula, or an
equation to represent the nth term of the sequence. Consider for example the sequence of odd numbers we started
with: 1,3,5,7,...
We can find an explicit formula for the nth term of the sequence if we analyze a few terms:
a1 = 1
a2 = a1 + 2 = 1 + 2 = 3
a3 = a2 + 2 = 1 + 2 + 2 = 5
a4 = a3 + 2 = 1 + 2 + 2 + 2 = 7
a5 = a4 + 2 = 1 + 2 + 2 + 2 + 2 = 9
a6 = a5 + 2 = 1 + 2 + 2 + 2 + 2 + 2 = 11
Note that every term is made up of a 1, and a set of 2’s. How many 2’s are in each term?
TABLE 1.9:
a1
a2
a3
a4
a5
a6
=1
=1+2=3
=1+2×2=5
=1+3×2=7
=1+4×2=9
= 1 + 5 × 2 = 11
7
1.2. Explicit Formulas
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The nth term has (n - 1) 2’s. For example, a99 = 1 + 98 × 2 = 197 . We can therefore represent the sequence as an =
1 + 2(n - 1). We can simplify this expression:
TABLE 1.10:
an
an
an
= 1 + 2(n - 1)
= 1 + 2n - 2
= 2n - 1
In general, we can represent an arithmetic sequence in this way, as long as we know the first term and the common
difference, d. Notice that in the previous example, the first term was 1, and the common difference, d, was 2. The
nth term is therefore the first term, plus d(n - 1):
TABLE 1.11:
an
=a1 + d(n - 1)
You can use this general equation to find an explicit formula for any term in an arithmetic sequence.
Example A
Find an explicit formula for the nth term of the sequence 3, 7, 11, 15... and use the equation to find the 50th term in
the sequence.
Solution
an = 4n - 1 , and a50 = 199
The first term of the sequence is 3, and the common difference is 4.
TABLE 1.12:
an
an
an
an
= a1 + d(n - 1)
= 3 + 4(n - 1)
= 3 + 4n - 4
= 4n - 1
a50
= 4(50) - 1 = 200 - 1 = 199
We can also find an explicit formula for a geometric sequence. Consider again the sequence in example 2a:
8
www.ck12.org
Chapter 1. Sequences, Series, and Mathematical Induction
TABLE 1.13:
t1 = 3
tn = 2 × tn−1
t2 = 2t1 = 2 × 3 = 6
t3 = 2t2 = 2 × 6 = 12
t4 = 2t3 = 2 × 12 = 24
t5 = 2t4 = 2 × 24 = 48
→
Notice that every term is the first term, multiplied by a power of 2. This is because 2 is the common ratio for the
sequence.
TABLE 1.14:
t1
t2
t3
t4
t5
=3
=2×3=6
= 2 × 2 × 6 = 22 × 6 = 12
= 2 × 2 × 2 × 6 = 23 × 6 = 24
= 2 × 2 × 2 × 2 × 6 = 24 × 6 = 48
The power of 2 in the nth term is (n-1). Therefore the nth term in this sequence can be defined as: tn = 3(2n −1 ). In
general, we can define the nth term of a geometric sequence in terms of its first term and its common ratio, r:
TABLE 1.15:
tn
= t1 (rn−1 )
You can use this general equation to find an explicit formula for any term in a geometric sequence.
Example B
Find an explicit formula for the nth term of the sequence 5, 15, 45, 135... and use the equation to find the 10th term
in the sequence.
Solution
an = 5 × 3n −1 , and a10 = 98,415
The first term in the sequence is 5, and r = 3.
TABLE 1.16:
an
an
a10
a10
= a1 × rn −1
= 5 × 3n−1
= 5 × 310−1
= 5 × 39 = 5 × 19,683 = 98,415
Again, it is always possible to write an explicit formula for terms of an arithmetic or geometric sequence. However,
you can also write an explicit formula for other sequences, as long as you can identify a pattern. To do this, you
9
1.2. Explicit Formulas
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must remember that a sequence is a function, which means there is a relationship between the input and the output.
That is, you must identify a pattern between the term and its index, or the term’s “place” in the sequence.
Example C
Write an explicit formula for the nth term of the sequence 1, (1/2), (1/3), (1/4)...
Solution
an = (1/n)
Initially you may see a pattern in the fractions, but you may also wonder about the first term. If you write 1 as (1/1),
then it should become clear that the nth term is (1/n).
TABLE 1.17:
Concept question wrap-up "They figure that memberships are increasing by about 500 people every three days.
At this rate, how many members will they have on the 48th day? How many days will it be before they reach
25,000 members?"
This is actually a fairly simple arithmetic sequence: each day there are 500/3 more members, on average. Use the
formula for arithmetic sequences from the first example above.
Vocabulary
An explicit formula for a sequence allows you to find the value of any term in the sequence.
The natural numbers are a subset of the integers: {1,2,3,4,5....}
A recursive formula for a sequence allows you to find the value of the nth term in the sequence if you know the
value of the(n-1)th term in the sequence.
A sequence is an ordered list of objects.
Guided Practice
Questions
1) Write an explicit formula for the sequence: 2, 9, 16... and use the formula to find the value of the 20th term.
2) Write an explicit formula for the sequence: 5, 10, 20... and use the formula to find the value of the 9th term.
3) Write an explicit formula for the sequence: (1/2), (1/4), (1/8) and use the formula to find the value of the 7th term.
4) Identify all sequences in the previous three problems that are geometric. What is the common ratio in each
sequence?
5) The membership of an online dating service increases at an average rate of 8% per year. In the first year, there are
500 members.
a. How many members are there in the second year?
b. How many members are there in the eighteenth year?
Solutions
1) For the sequence: 2, 9, 16...
an = 7n − 5
10
www.ck12.org
Chapter 1. Sequences, Series, and Mathematical Induction
∴ a20 = 7(20) − 5
a20 = 135
2) For the sequence: 5, 10, 20...
an = 5 · 2n−1
∴ a9 = 5 · 28
a9 = 5 · 256
a9 = 1280
3) For the sequence: (1/2), (1/4), (1/8)...
an =
1
2n
∴ a7 = 217
1
a7 = 128
4) The sequence in question 1 is arithmetic.
The sequence is question 2 is geometric, and has r = 2.
The sequence in question 3 is geometric and has r = 1/2.
5) a. 540 members
b. Approximately 1,998 members
Practice
Name the sequence as arithmetic, geometric, or neither.
1.
2.
3.
4.
−21, −6, 18, −3, 20, −2
−2 −3 −4
0, −1
5 , 5 , 5 , 5 , −1
1, 3, 9, 27, 81, 243
2, 9, −2, 1, 18, 2
Write the first 5 terms of the arithmetic sequence(explicit).
5. an = −8 − 9(n − 1)
6. an = 6 − 23 (n − 1)
7. an = 8 + 13 (n − 1)
Solve the following:
8.
9.
10.
11.
12.
13.
What are the first five terms of the sequence? an = an−1 − 10
3 ; a1 = −6
Given the sequence, write a recursive function to generate it: 2, −4, −10, −16, −22, −28
Write the equation of an without using recursion: an = an−1 − 32 ; a1 = 10
Write as a recursion: an = 6 − 53 (n − 1)
Write the equation of an without using recursion: an = an−1 + 8; a1 = 3
What are the first five terms of the sequence? an = an−1 − 1; a1 = −5
11
1.2. Explicit Formulas
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Write the formula for the explicit(nth )term of the arithmetic sequence.
14.
15.
16.
17.
−5
11 19
−7, −13
3 , 3 , 1, 3 , 3
6, −4, −14, −24, −34, −44
9, 16, 23, 30, 37, 44
In a particular arithmetic sequence, the second term is 4 and the fifth term is 13. Write an explicit formula for
this sequence.
Write the first 5 terms of the geometric sequence
17. an = 5(−3)(n−1)
(n−1)
18. an = −6( −10
)
3
Write the formula for the nth term of the geometric sequence
19. −8, 16, −32, 64, −128, 256
81 243 729 2187
20. 9, 27
2 , 4 , 8 , 16 , 32
Convert the explicit and rewrite as recursion:
(n−1)
21. an = 9( −4
3 )
22. an = −6(−4)(n−1)
23. an = −5(5)(n−1)
12
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Chapter 1. Sequences, Series, and Mathematical Induction
1.3 Sum Notation and Properties of Sigma
Here you will learn how to read basic summation notation, and you will explore a few basic operations using sums
and sigma notation.
Sayber and Tuscany sell popsicles during the summer for pocket money. One particular weekend, they purchased a
package of 30 popsicles from the store.
Usually they just offer the popsicles for free, 1 per customer, and accept tips. This time, Tuscany wonders if they
would make more money by charging $0.50 per popsicle. At the same time, Sayber wonders if he might be able to
increase his tips by encouraging customers to "outbid" each other. The two children decide to each take 15 popsicles
and see who makes the most.
How could you calculate how much money each of them makes, assuming Sayber gets a $0.10 tip from the first
customer, and is able to convince each successive customer to double the previous person’s tip?
Watch This
Embedded Video:
MEDIA
Click image to the left for more content.
- James Sousa: Sigma Notation
Guidance
Consider for example a sequence defined by an = 3n. If we write out the sum of the first 4 terms, we have 3 + 6 + 9
+ 12 = 30. But what if we want to write out terms for a larger sum?
Summation notation is a method of writing sums in a succinct form. To write the sum 3 + 6 + 9 + 12 = 30 , we use
the Greek letter Sigma, as follows:
TABLE 1.18:
∑4n=1 3n
The expression 3n is called the summand, the 1 and the 4 are referred to as the limits of the summation, and the n is
called the index of the sum. Here we have used a “sigma” to write a sum. We can also read a sigma, and determine
the sum. For example, we can read the above sigma notation as “find the sum of the first four terms of the series,
where the nth term is 3n.” We always read the limits from the bottom to the top. The bottom number tells you which
term to start with, and the top limit tells you which term is the final term to add. We could then write out the sigma
above as:
13
1.3. Sum Notation and Properties of Sigma
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TABLE 1.19:
∑4n=1 3n
= 3(1) + 3(2) + 3(3) + 3(4)
= 3 + 6 + 9 + 12 = 30
In general, we can either rewrite a given series in sigma notation, or we can read sigma notation in order to find the
value of the sum.
Properties of Sigma
Notice that we can write the sum 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 19 + 20 as 2(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9
10
+ 10) Therefore ∑10
n=1 2n = 2 ∑n=1 n. In general, we can factor a coefficient out of a sum:
TABLE 1.20:
∑kn=1 can
= c ∑kn=1 an
Example A
Write the sum using sigma notation:
2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 19 + 20
Solution
∑10
n=1 2n
Every term is a multiple of 2. The first term is 2 × 1, the second term is 2 × 2 , and so on. So the summand of the
sigma is 2n. There are 10 terms in the sum. Therefore the limits of the sum are 1 and 10.
Example B
Write out the terms of ∑5n=2 (n + 7) and evaluate the sum.
Solution
Break the sum into two different sums. The sum is 37.
TABLE 1.21:
∑5n=2 (n + 7)
=(2 + 7) + (3 + 7) + (4 +
7) + (5 + 7)
=2 + 3 + 4 + 5 + 7 + 7 + 7
+7
=2 + 3 + 4 + 5 + 7 × 4 =
14 + 28 = 42
Notice we could have written ∑5n=2 (n + 7) as ∑5n=2 n + ∑5n=2 7 . Also, the second sum does not depend on the index
of the sum (i.e. it stays at 7 regardless of the index), only that there are 4 terms to add together. Seeing this can make
a sum easier to evaluate.
Example C
Write out the terms of ∑4n=0 32
14
1 n
4
and evaluate the sum.
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Chapter 1. Sequences, Series, and Mathematical Induction
Solution In general, we can write a sum as a sum of sums: ∑nn=1 (an + bn ) = ∑nn=1 (an ) + ∑nn=1 (bn ).
The sum is 42 85
TABLE 1.22:
∑4i=0 32
1 n
4
0
1
= 32 14 + 32 14 +
2
3
32 14
+ 32 14
+
1 4
32 4
1
= 32 · 1 + 32 · 14 + 32 · 16
+
1
1
32 · 64 + 32 · 256
= 32 + 8 + 2 + 12 + 18 =
42 58
TABLE 1.23:
Concept question wrap-up
Tuscany’s income can be expressed as: ∑10
n=1 .5 → .5 · 10 = $5.00
10
Sayber’s income can be expressed as: ∑n=1 .10 · 2n → .10(20 ) + .10(21 ) + .10(22 )... + .10(210 ) → $102.30
Sayber must be a heck of a salesman to get that last tip!
Vocabulary
Sigma is the large, stylized capitol E that means, approximately, "the sum of".
The index of the sum is the variable in the sum.
The summand describes how to manipulate each term in a series.
The limits of the sum are written above and below the sigma, and describe the domain to be used in the series
calculation.
Guided Practice
Questions
1) 2 + 6 + 18 + 54
∑4n=1 2 3n−1 or ∑3n=0 2 (3n )
2) Find the series of numbers and the total of those numbers of the arithmetic series represented by the following
3
sigma notation: ∑14
n=8 3 + 4 (n − 1)
3) Find the sum of all the numbers in the arithmetic sequence. ∑28
n=8 9 − 2(n − 1)
4) Find the series of numbers and the total of those numbers of the geometric series represented by the sigma notation
n−1
∑8n=1 7( −1
2 )
n−1
5) Find the sum of the terms in: ∑11
n=1 9(4)
Solutions
1) ∑5n=1 2n − 1
15
1.3. Sum Notation and Properties of Sigma
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2) To find the series of numbers, we plug in all the numbers between 8 and 14 for (n)
a) 3 + 34 ((8) − 1) = 33
4
b) 3 + 43 ((9) − 1) = 9
We would continue to do this clear through the number 14. Leaving us with the following series:
51
45
4 + 12 + 4
33
4
21
+ 9 + 39
4 + 2 +
This is fine, if we are just looking for the individual numbers in the sequence, however when asked to evaluate
sumations we are being asked to add all the numbers of the series together. It took plenty long enough to find
each number, and now we must add them all together. Fortunately there is a formula, not only eliminating our
need to find each number, but that lets us also add them all together and arrive at our answer or sum.
The formula is 2k (a1 + an ) works like this:
We plug in n = 8 to get a8 = 33
4
Then we plug in n = 14, to get a14 = 51
4
Identify the number of terms (k) = 14 − 8 + 1 so we use k = 7 in the formula below.
51
147
Now we can use the formula: 2k (a8 + a14 ) and we get: 27 ( 33
4 + 4 )= 2
∴ ∑4n=1 2 3n−1 or ∑3n=0 2 (3n )
3) Use the formula 2k (a1 + an ) to find the sum of ∑28
n=8 9 − 2(n − 1)
Substitute n = 8 to get a8 = −5
Substitute n = 28 to get a2 8 = −45
Identify the number of terms as: k = 28 − 8 + 1 so we use k = 21 in our formula below:
Now we can use the formula: 2k (a8 + a14 ) and our answer is −525
∴ ∑10
n=1
1
n
n−1
4) To identify and sum the terms in the geometric series ∑8n=1 7( −1
2 )
a) Find the sequence of numbers the same way as we did in problem 2, by plugging in the indicated numbers 1-8
for (n)
8
((1)−1) which gives us: 7
b) ∑n=1 7( −1
2 )
7
−7
7
−7
7
−7
c) We do this for the remaining numbers and our sequence looks like this:7 + −7
2 + 4 + 8 + 16 + 32 + 64 + 128
As in prior examples, we are not just being asked to find the numbers, but add them all together. Again, we do not
want to have to take the time to find all the numbers and add them all together. Once again we find ourselves lucky,
there is a formula!
k
The formula a1 ( 1−r
1−r ) works like this:
We plug n=1 into a1 which gives us 7.
k = 8, and r = −1
2
1− −1
8
2
Substituting our numbers results:7( 1− −1
)
Which gives us
595
128
2
5) Let’s use the formula for geometric series again:
Identify the terms: a1 = 9, k = 11 and r = 4
11
Substituting into our formula, we have: 9( 1−4
1−4 ) = 12, 582, 909
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Chapter 1. Sequences, Series, and Mathematical Induction
Practice
Express the Sum using Sigma Notation:
1. 1 + 3 + 5 + 7 + 9
1
2. 1 + 12 + 13 + 14 + ... + 10
Find the series of numbers indicated by the sigma notation:
10
3. ∑14
n=0 −2 − 3 (n − 1)
14
4. ∑n=−8 −7 − 3(n − 1)
Evaluating Summations of an Arithmetic Series
5. ∑5n=−10 7 − 43 (n − 1)
6. ∑3n=−3 3 − 31 (n − 1)
7. ∑1n=−5 −6 + 34 (n − 1)
Find the series of numbers indicated by the sigma notation
10. ∑2n=1 3(− 21 )n−1
11. ∑5n=1 5(− 34 )n−1
Evaluating Summations of Geometric Series
12.
13.
14.
15.
16.
∑6n=1 3( 21 )n−1
∑7n=1 −5(− 21 )n−1
4 n−1
∑11
n=1 −7(− 3 )
∑6n=1 −7( 41 )n−1
∑3n=1 2(− 23 )n−1
17
1.4. Series Sums and Gauss’ Formula
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1.4 Series Sums and Gauss’ Formula
Here you will learn of a formula that makes it easier to find the sum of the terms in a series, and will further explore
the use of summation notation and number series.
EyeScreem Ltd is having a special sale on their ice cream cones. In order to get as many people in the store to try
out their new flavors, they have decided to run the following promotion:
The first customer to buy an ice cream will pay $6 for a large cone with chocolate fudge.
The second customer will pay only $5.90.
The third customer will only pay $5.80.
Each successive ice cream lover will be charged $0.10 less than the last, until everyone that comes in gets a free
cone!
How much money will the store bring in during the sale, assuming at least some customers are given free cones?
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- PatrickJMT : Introduction to Arithmetic Series Sums
Guidance
Gauss’ Formula
German mathematician C.F. Gauss is often credited with discovering a formula for calculating the sum of a series
when he was a young child. The story is likely apocryphal (a legend), but it has been passed down since Gauss lived
in the 1700’s. According to the story, Gauss’s teacher wanted to occupy students by having them add up large sets
of numbers. When Gauss was asked to add up the first 100 integers, he found the sum very quickly, by pairing the
numbers:
All of the numbers in the sum could be paired to make groups of 101. There are one hundred numbers being added,
so there are such fifty pairs. Therefore the sum is 50(101) = 5050.
18
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Chapter 1. Sequences, Series, and Mathematical Induction
The method Gauss used to solve this problem is the basis for a formula that allows us to add together the first n
positive integers:
∑=
(n)(n+1)
2
Using a graphing calculator
To generate either a sequence or a series, you can use a graphing calculator. The TI-83/84 series gives you several
options. You can, for example, work in sequence mode, which allows you to define a sequence and find terms. If you
have an explicit formula for a sequence, you can keep your calculator in function mode. For example, consider the
sum ∑9n=1 n2 . It would be time consuming to write out the first 9 squares. The calculator is faster. To generate the
9 terms, press <TI font_2nd>[LIST], then select OPS, then option 5, seq(. This takes you back to the main screen.
You should see seq(. After this, enter x^2, x, 1, 9, 1). (The x tells the calculator that x is the input. The 1 and the 9
tell it the limits of the sum. The second 1 tells the calculator to go up in increments of 1.)
Press <TI font_ENTER>, and you should see the list of squares. Scroll to the right to see all of them. The scrolling
will end when you reach 81.
If you want to find the sum of the terms, first store the sequence in a list (see screen below), then <TI font_2nd>[LIST], then select MATH, then option 5, sum(.Then enter the name of the list and press <TI font_)><TI
font_ENTER>. You should get 285.
Example A
Expand the sigma and find the sum. ∑7n=1 (2n − 3)
Solution
TABLE 1.24:
∑7n=1 (2n − 3)
= (2 × 1 − 3) + (2 × 2 − 3) + (2 × 3 − 3) + (2 × 4 − 3) +
(2 × 5 − 3) + (2 × 6 − 3) + (2 × 7 − 3)
= (−1) + (1) + (3) + (5) + (7) + (9) + (11)
= 35
Example B
Expand the sigma and find the sum by adding the terms ∑6n=3 (n2 − 5)
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1.4. Series Sums and Gauss’ Formula
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Why is Gauss’ formula not recommended for this question?
Solution
TABLE 1.25:
∑6n=3 (n2 − 5)
= (32 − 5) + (42 − 5) + (52 − 5) + (62 − 5)
= (4) + (11) + (20) + (31)
= 66
There are actually a couple of reasons not to use Gauss formula here, but the biggest is that the formula assumes you
are adding all of the integers from 0 to the last number in the series. In the question, you are asked to only sum from
the 3rd to 6th term.
Example C
If the sum of the first n integers is 210, what is n?
Solution
Using Gauss’ formula:
∑=
(n)(n+1)
2
210 = (n)(n+1)
2
Multiply both sides by 2 420 = (n)(n + 1)
Distribute 420 = (n2 + n)
Complete the square 420 = (n2 + n + 1/4)
Factor 420 = (n + 1/2)2
Square root both sides 201/2 = n + 1/2
n = 20
TABLE 1.26:
Concept question wrap-up
The series sum formula ∑ = (n)(n+1)
is designed for integers, so let’s use it to solve for the number of dimes
2
brought in (since that is the unit each term reduces by) and then convert to dollars:
(60)(61)
∑= 2
3,660
∑= 2
∑ = 1830.00 → $183.00
The store will bring in $183.00, which will probably not cover the costs of the day. However, they will certainly
get a lot of people through the door to try out the ice cream!
Vocabulary
If a series has a limit, and the limit exists, the series converges.
If a series does not have a limit, or the limit is infinity, then the series diverges.
20
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Chapter 1. Sequences, Series, and Mathematical Induction
The index of the sum is the variable in the sum.
The limits of a summation are the starting and ending points of the sum.
Sigma Σ is the Greek letter used to represent a sum.
The summand of a sigma is the expression being summed.
Guided Practice
Questions
1) Express the sum using sigma notation: 1 + 3 + 9 + 27 + ...
−11
2) Find the sum of ∑12
n=0 3 (n − 1)
3) Use Gauss’s formula to find the sum of the first 200 positive integers.
1
4) Expand the sigma and find the sum. ∑5n=1 3n
Solutions
n−1 or ∞ 3n
1) ∑∞
∑n=0
n=1 3
2) Let’s look at both ways of solving this one:
−11
−13
−17
−19
a) We could plug in all the numbers between 0 and 12 to get: −7
3 + (−3) + 3 + 3 + (−5) + 3 + 3 + (−7) +
−23
−25
−29
−31
3 + 3 + (−9) + 3 + 3 ... and then add them together to get the sum
b) Using a formula:
A slightly modified version of Gauss’ formula looks this: Sn = 2k (a0 + an ) where k is the number of terms in the
series plus one, and a0 and an are the first and last terms in the series
13
2
−7
3
+ −31
=
3
−247
3
→ −82 13
3) Using Gauss’ formula:
=
(n)(n+1)
2
n = 200
=
(200)(200+1)
2
Sum = 20,100
4) For ∑5n=1
1
3n
TABLE 1.27:
∑5n=1
1
3n
=
=
=
1
3
60
180
137
180
1
+ 16 + 19 + 12
+
30
20
15
12
+ 180
+ 180
+ 180
+ 180
1
15
Practice
Calculate the sums of the given series, you may use addition of individual terms, or a series sum formula. You may
use a graphing tool for any 3 of them. Try to use each method at least once.
1. ∑16
n=0 −10 + 3(n − 1)
21
1.4. Series Sums and Gauss’ Formula
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
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64 + 72 + 80 + ... + 200
17
17
−15
4 − 4 − 4 + ... 2
6
1
∑n=0 6 − 2 (n − 1)
−71
67
37
3 − 3 − 21 + ... + 3
2 + 4 + 6 + ...26
−2 − 1 + 0 + ... + 12
13
31
71
− 21
4 − 2 − 4 + ... − 4
43
− 72 − 8 − 25
2 + ... − 2
20
∑n=7 −1 + (n − 1)
∑22
n=−6 5 + 3(n − 1)
3
∑13
n=−2 −5 − 2 (n − 1)
Consider the sums ∑5n=1 (n + 1) and ∑5n=1 (n − 4).
13. What is the product of ∑5n=1 (n + 1) ∑5n=1 (n − 4)
14. What is the sum of ∑5n=1 (n + 1)(n − 4)?
15. Look closer at the last two problems, what does this tell you about rules for working with sums?
22
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Chapter 1. Sequences, Series, and Mathematical Induction
1.5 Problem Solving with Series Sums
Here you will learn more about how to calculate the sum total of all the members of arithmetic series, given the
starting and ending numbers, and the difference between each term.
Katrina just started her new job last week. She is making $60 per day, and saving 10%. Her savings account has
been slowly building up at $6 per day, and currently has $73 in it.
How much money will she have after working another 15 days? How much after another 27 days?
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- James Sousa: Arithmetic Series
Guidance
In the previous lesson, we discussed Gauss’ formula for finding the total of the numbers in a series. In this lesson,
we continue to practice that, and also to make use of the modified version of Gauss’ Formula that is commonly seen
in other texts:
The sum of the first n terms in an arithmetic series is Sn =
n(a1 +an )
2
Since this lesson is mostly about applying the rules for series, there is minimal instruction. If you would like a more
in-depth presentation on series, please see the prior lesson: "Series Sums and Gauss’ Formula."
Example A
Find the sum of the first 50 terms of an arithmetic series if the first term is 5 and the common difference is 3.
Solution
50(a1 +an )
..... Substitute the applicable number of terms in the
2
50(5+(5+49×3))
..... Substitute the values for the first and 50th terms
2
50(157)
= (25)(127) ..... Simplify
2
S50 =
sequence
3, 175
This method is clearly much easier than writing out and adding 50 numbers!
23
1.5. Problem Solving with Series Sums
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Example B
Find the sum of the first 40 terms of an arithmetic series in which the first term is 8 and the common difference is 5.
Solution
Use the formula Sn =
n(a1 +an )
2
..... Substitute in the number of terms, and the value of the first and last terms
S40 = 40(8+39·5)
2
40(8+195)
S40 =
..... Simplify
2
S40 = 8120
.....
Simplify
2
S40 = 4060
Example C
Consider the series: -13 + -3 + 7 + 17 + 27 + ...
a) What is the 25th term?
b) What is the sum of the first 25 terms?
Solution
a) This is an arithmetic series where each successive term is 10 more than the last
−13 + 24(10) = 227
b) Use the formula Sn = n(a12+an )
This is the sum of the first 25 terms of an arithmetic series with a common difference of 10
S25 = 25(−13+227)
..... Substitute in the number of terms, and the value of the first and last terms
2
S25 = 25·214
.....
Simplify
2
S25 = 5350
2 ..... Simplify
S25 = 2675
Vocabulary
A hypothesis is a conclusion made on the basis of evidence, or a statement assumed to be true for the sake of an
argument.
Mathematical induction is a method of mathematical proof used to establish that a given statement is true of all
positive integers (natural numbers).
A partial sum is the sum of the first n terms in an infinite series, where n is some positive integer.
Guided Practice
Questions
1) A student needed to find the sum of the first 10 terms of the series 4 + 12 + 36 + ... and so he wrote the following:
9
)
S10 = 10(4+4×3
= 10(78,736)
= 393, 680
2
2
Do you agree with the student’s work? Explain.
2)Given an arithmetic sequence (an ) determined by a1 = 143 and d = −3
24
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Chapter 1. Sequences, Series, and Mathematical Induction
What is the 220th number of the sequence.
We find the solution by using the following formula:
an = a + (n − 1)d
3)How do we find the sum of the 220th number of the sequence, given the same information as the 2nd guided
practice above?
We use the following formula: sn =
a1 +an
2 (n).
We set n = 220.
Solutions
1) Because the series is geometric, this formula is not appropriate. The work here does not represent the sum of the
first 10 terms. Using a graphing calculator, you can find that the sum is 118,096.
2) To find the 220th number:
a220 = a1 + 219d
= 143 + 219(−3)
= −514
3) To find the sum of the first 220 numbers:
220
(220)
s220 = a1 +a
2
143+(−514)
s220 =
(220)
2
s2 20 = −371
·
220
2
the 220th number of the sequence is = −40, 810
Practice
Use the arithmetic formula to solve the following problems:
1. Given the arithmetic series of numbers: 1, 4, 7, 10, 13... a)Find the 200th number in the sequence b)find the
sum of the first 200 numbers.
2. The sum of the first five numbers of an arithmetic sequence is -45. What is the value of the third number? (
hint: find a3 if s5 = −45)
3. You have an arithmetic series of numbers defined by: a1 = 45 and d = −5 a)Determine a1 50 b)Identify the
sum of: a1 + a2 + ... + a150
4. The sum of the first three numbers in an arithmetic sequence is 219. The sum of the first nine numbers in the
same sequence is 603. What is the 143rd number of the sequence.
5. The first eight numbers of an arithmetic sequence add up to 604. The next eight numbers added up equal 156.
Find the first number and the common difference in the sequence.
6. The first number in an arithmetic sequence is 80. Find the common difference if we also know that s9 is
eighteen times a11
7. If an is an arithmetic sequence with a1 = 1. Find the second number if we know that the sum of the first five
numbers is one-fourth of the sum of the next five numbers.
8. Given (an ) = 78, 75, 72, 69...Finda150 and s150
9. The following conditions exist within a sequence of numbers: a50 = 252 and s50 = 2800 What is the first
number of the series, and what is the common difference?
10. What are the values of a and d, given that (an ) is an arithmetic series of numbers, if we know: a15 = 62 and
s20 = 700?
11. Given the sequence: a1 = −16 and d = 31 Find the values of n, so that sn = 50
25
1.5. Problem Solving with Series Sums
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12. If a1 = 8 and d = −3 What are the values of a20 and s20
13. If a34 = 193 and s17 = 306 find a and d
14. Three sides of a right triangle are integers and form an arithmetic sequence. Find the sides of the triangle
26
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Chapter 1. Sequences, Series, and Mathematical Induction
1.6 Inductive Proofs
Here you will learn about proof by induction, a process which allows you to attempt to prove a theorem is true for
every number if it is true for a single particular number.
Proving a theory can be a daunting process, after all, no matter how many times you try something with the same
result, how can you be certain that it will always have the same result, no matter what?
For example, if you were to see someone fill a water balloon with ice water and hold it out the window, you
would probably either cringe in anticipation of the shouting below, or eagerly watch, depending on the situation. In
either case, your response would be based on the fact that you would be certain that a water balloon would pop on
someone’s head if dropped out the window onto them. Your certainty would be based on your past experience with
water balloons and sidewalks, and you’d very very likely be correct, but until the balloon actually hits the target,
there isn’t any way to be absolutely certain it will break.
In math, situations like this occur a lot. Based on repeated experience, you may develop a rule or shortcut to save
time or effort when calculating. However, you may be rightly concerned about using such shortcuts on an important
exam. After all, how can you be certain that the shortcut works in every situation?
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- PatrickJMT: Proof by Induction
Guidance
In this lesson you will learn about mathematical induction, a method of proof that will allow you to prove that a
particular statement is true for all positive integers.
Inductive Proofs
First let’s make a guess at a formula that will give us the sum of all the positive integers from 1 to n for any integer
n. If we look closely at Gauss’s Formula we used in the last lesson, where the young boy was able to quickly add
up all of the numbers between 1 and 100, we can see a general form: there were 100 numbers, hence 50 pairs. So if
there were n numbers, there would be (n/2) pairs. The first and last numbers were 1 and 100. They added together
to give us 101. This number was the sum of each pair in the overall sum. So in general, we could add together 1 and
n to get the sum of each pair. Therefore we might hypothesize that the sum of the first n positive integers is n((1 +
n)/2). However, we have not proven that this formula works for all positive integers n. Mathematical induction will
allow us to do this.
The overall idea of induction is this: Assume that a statement is true for some arbitrary value of n, and show that if
the statement is true for n = k, it must also be true for n = k+1. This process is used because we can’t actually show
27
1.6. Inductive Proofs
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it is true for every value. For example, you might show that the above equation is true for n = 100, and then n = 101,
and then n = 102, but then what about 103? 104? 500? A million?
Mathematical induction allows us prove that a statement is true in three steps:
Step 1) The base case: prove that the statement is true for the first value of n. In some cases, this might be n = 0. In
the case of the integer sum formula above, we would start with n = 1. Often with induction you may want to expand
the first step by showing that the statement is true for several values of n.
Step 2) The inductive hypothesis: assume that the statement is true for the kth value of n. In the case of the integer
sum formula, we would state the following: the sum of the first k positive integers is k((1 + k)/2).
Step 3) The inductive step: use the inductive hypothesis to show that the statement is true for the k + 1th step. In the
case of the integer sum formula, we would prove the following: assuming that the sum of the first k positive integers
is k((1 + k)/2) , the sum of the first k +1 positive integers is ((k + 1) (1 + (k + 1)/2) .
Carrying out this kind of proof requires that you perform each of these steps., For the third step in particular you
must rely on your algebra skills.
Example A
Prove: The sum of the first n positive integers is
n(1+n)
2
Solution
Use the three steps of mathematical induction:
Step 1) The base case:
If n = 1, the entire sequence is just 1 and therefore the sum is 1. Also,
This establishes the base case.
Step 2) Assume that the sum of the first k positive integers is
In other words, assume that 1 + 2 + 3 + · · · + k =
k(1+k)
2
n(1+n)
2
=
1(1+1)
2
=
2
2
= 1.
.
k(1+k)
2 .
Step 3) We must show that the sum of the first k + 1 positive integers is
In other words, we must show that 1 + 2 + 3 + · · · + k + (k + 1) =
(k+1)(1+(k+1))
.
2
(k+1)(1+(k+1))
.
2
There are two key ideas to keep in mind as you are carrying out this step: (1) remember to use the assumption and
(2) remember how sums work.
How does the sum of the first k + 1 integers relate to the sum of the first k integers? To get the sum of the first k +
1 integers we must add up all the integers from 1 to k and then add on k + 1, since the sum of the first k + 1
integers is 1 + 2 + 3 + · · · + k + (k + 1).
Now we must use our assumption. Remember that we are assuming that 1 + 2 + 3 + · · · + k = k(1+k)
2 .
k(1+k)
Substitute 2 in for 1 + 2 + 3 + · · · + k in our expression above for the sum of the first k + 1 integers.
Now we have the sum of the first k + 1 integers is k(1+k)
+ (k + 1).
2
Remember that we are trying to show that the sum of the first k + 1 integers is
manipulation, we can show that k(1+k)
+ (k + 1) = (k+1)(1+(k+1))
.
2
2
See below:
28
(k+1)(1+(k+1))
.
2
With some algebraic
www.ck12.org
Chapter 1. Sequences, Series, and Mathematical Induction
TABLE 1.28:
k(1+k)
+ (k + 1)
2
k(k+1)
= 2 + 2(k+1)
2
The common denominator is 2
Add the fractions
=
=
=
=
=
k(k+1)+2(k+1)
2
k2 +k+2k+2
2
k2 +3k+2
2
(k+1)(k+2)
2
(k+1)((k+1)+1)
2
Simplify the numerator
Factor the numerator
The term (k + 2) is the same as ((k + 1) + 1)
We have shown that our formula for the sum of the first n integers is true for n = 1. We have also shown that whenever
it is true for n = k it is also true for n = k+1. Since we know it is true for n = 1, it must therefore be true for n = 2.
Similarly, since it is true for n = 2, it must therefore be true for n = 3, and it must therefore be true for n = 4,... You
should see that we have proven that the sum of the first n positive integers is n(1+n)
for all integer values of n. We
2
can similarly prove a formula for the sum of the first n terms in an arithmetic series.
Example B
Prove that the sum of the first n terms of an arithmetic series is Sn =
and an is the last term.
n(a1 +an )
2
where a1 is the first term in the series
Solution
Recall that in an arithmetic sequence or series, there is a common difference, d, between each term, and that the nth
term is an = a1 + d(n − 1) We need to keep these ideas in mind in order to complete the proof.
Step 1) Base case: if n = 1, then S1 = a1
Using the hypothesized formula, we have
TABLE 1.29:
S1 =
Step 2) Assume that Sk =
1(a1 +a1 )
2
=
2a1
2
= a1
k(a1 +ak )
2
Step 3) Prove that if our formula for Sk is true then Sk+1 =
(k+1)(a1 +ak+1 )
.
2
We can think of the sum of the first k + 1 terms as the sum of the first k terms, plus the k + 1 term. So we have:
TABLE 1.30:
Sk+1 = Sk + ak+1
= k(a12+ak ) + ak+1
= k(a12+ak ) + 2a2k+1
= k(a1 +ak2)+2ak+1
Add the k + 1 term
Use the formula for Sk from step 2
The common denominator is 2
Add the fractions
29
1.6. Inductive Proofs
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TABLE 1.30: (continued)
Sk+1 = Sk + ak+1
1 +kd)
= k(a1 +(a1 +(k−1)d))+2(a
2
=
=
=
=
=
=
=
Add the k + 1 term
Use substitution: remember that am = a1 + (m − 1)d
for any positive integer m. So ak = a1 + (k − 1)d and
ak+1 = a1 + (k)d
k(a1 +a1 +kd−d)+2a1 +2kd
2
ka1 +ka1 +k2 d−kd+2a1 +2kd
2
2ka1 +2a1 +k2 d+kd
2
2a1 (k+1)+kd(k+1)
2
(k+1)(2a1 +kd)
2
(k+1)(a1 +a1 +kd)
2
(k+1)(a1 +ak+1 )
2
Distribute and combine like terms
Factor by grouping
Again, ak+1 = a1 + (k)d
Example C
Use induction to prove that 12 + 22 + 32 + · · · + n2 =
n(n+1)(2n+1)
6
Solution
Step 1) Base case: 12 = 1
1(1+1)(2(1)+1)
6
=
2(3)
6
=1
Step 2) Inductive hypothesis: 12 + 22 + 32 + ... + k2 =
k(k+1)(2k+1)
6
Step 3) Inductive step: show that
12 + 22 + 32 + ... + k2 + (k + 1)2 =
First, note that
Now we have:
(k+1)(k+1+1)(2(k+1)+1)
6
(k+1)(k+1+1)(2(k+1)+1)
6
=
(k+1)(k+2)(2k+3)
.
6
TABLE 1.31:
12 + 22 + 32 + ... + k2 + (k + 1)2
= k(k+1)(2k+1)
+ (k + 1)2
6
2
= k(k+1)(2k+1)+6(k+1)
= (k+1)[k(2k+1)+6(k+1)]
6
6
(k+1)[2k2 +k+6k+6]
(k+1)[2k2 +7k+6]
=
=
= (k+1)(2k+3)(k+2)
6
6
6
TABLE 1.32:
Concept question wrap-up
Situations like this are custom-made for inductive proofs. If you come up with a shortcut on your math, and want
to be absolutely certain it works in every situation, run it through the proof explained in this lesson. If it passes
all of the tests, you can be sure it will work with any number you throw at it.
30
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Chapter 1. Sequences, Series, and Mathematical Induction
Vocabulary
Mathematical induction, allows you to prove that a particular statement is true for all positive integers by proving it
true for "n", and "n + 1".
A series sum is the total of all of the numbers in a series.
The nth term in a series commonly refers to the last term in a series, often left unspecified.
Guided Practice
Questions
1) Use induction to prove that 1 + 3 + 5 + · · · + (2n − 1) = n2
2) Use induction to prove that 13 + 23 + 33 + · · · + n3 =
n2 (n+1)2
4
3) Use induction to prove that 1 + 4 + 7 + · · · + (3n − 2) =
3n2 −n
2
Solutions
1) Prove that 1 + 3 + 5 + · · · + (2n − 1) = n2
Step 1) Base case: 1 = 12
Step 2) Inductive hypothesis: assume that 1 + 3 + 5 + ... + (2k − 1) = k2
Step 3) Show that 1 + 3 + 5 + ... + (2k − 1) + (2k + 1) = (k + 1)2
We have: 1 + 3 + 5 + ... + (2k − 1) + (2k + 1) = k2 + (2k + 1)
TABLE 1.33:
= k2 + 2k + 1
= (k + 1)(k + 1) = (k + 1)2
2) Prove that 13 + 23 + 33 + · · · + n3 =
n2 (n+1)2
4
Step 1) Base case: 13 = 1
12 (1+1)2
4
=
22
4
=1
Step 2) Assume that 13 + 23 + 33 + ... + k3 =
k2 (k+1)2
4
Step 3) Show that 13 + 23 + 33 + ... + k3 + (k + 1)3 =
First, note that
(k+1)2 ((k+1)+1)2
4
=
(k+1)2 ((k+1)+1)2
4
(k+1)2 (k+2)2
4
Now we have:
TABLE 1.34:
13 + 23 + 33 + ... + k3 + (k + 1)3
2
2
= k (k+1)
+ (k + 1)3
4
2
2
3
= k (k+1) 4+4(k+1)
(k+1)2 [k2 +4(k+1)]
(k+1)2 [k2 +4k+4]
=
=
=
4
4
(k+1)2 (x+2)2
4
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1.6. Inductive Proofs
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3) Prove that 1 + 4 + 7 + · · · + (3n − 2) =
3n2 −n
2
Step 1) Base case: 1 = 1
3(1)2 −1
2
=
2
2
=1
Step 2) Inductive hypothesis: assume that 1 + 4 + 7 + ... + (3k − 2) =
Step 3) Show that 1 + 4 + 7 + ... + (3(k + 1) − 2) =
3k2 −k
2
3(k+1)2 −(k+1)
2
First note that:
TABLE 1.35:
1 + 4 + 7 + ... + (3(k + 1) − 2) =
= (k+1)[3(k+1)−1]
2
3(k+1)2 −(k+1)
2
(k+1)[3k+2]
2
Now we have:
TABLE 1.36:
1 + 4 + 7 + ... + (3k − 2) + (3(k + 1) − 2)
2
= 3k 2−k + (3(k + 1) − 2)
2
= 3k 2−k + (3k + 1)
2
2
3k2 −k+2(3k+1)
= 3k −k+6k+2
= 3k +5k+2
=
2
2
2
Practice
Use induction to prove the following:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
−15 − 25 − 35 + ... − 10k − 5 = k(−5k − 10)
k+1
1 + 9 + 92 + 93 + ... + 9k = 9 9−1−1
4 + 8 + 12 + ... + 4k = 2k(k + 1)
6 + 8 + 10 + ... + 2k + 4 = k(k + 5)
1 + 2 + 3 + ... + k = 21 k(k + 1)
k+1
1 + 6 + 62 + 63 + ... + 6k = 6 6−1−1
−1 − 5 − 9 + ... − 4k + 3 = k(−2k + 1)
k+1
1 + 4 + 42 + 43 + ... + 4k = 4 4−1−1
3 + 6 + 9 + ... + 3k = 32 k(k + 1)
−1 − 3 − 5 + ... − 2k + 1 = k(−k)
k+1
1 + 3 + 32 + 33 + ... + 3k = 3 3−1−1
n+1
12. 1 + 7 + 72 + 73 + ... + 7n = 7 6 −1
13. 4 + 8 + 12 + ... + 4n = 2n(n + 1)
32
(3k+2)(k+1)
2
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Chapter 1. Sequences, Series, and Mathematical Induction
14. 10 + 18 + 26 + ...8n + 2 = n(4n + 6)
15. 6 + 12 + 18 + ... + 6n = 3n(n + 1)
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1.7. Induction and Factors
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1.7 Induction and Factors
Here you will explore applications of mathematical induction, using it to prove basic properties of factors.
Proof by induction is common in mathematics, in this lesson we will use it to prove different kinds of hypothesis and
to practice the use of the process in different situations. Specifically, we will be applying the principles of induction
to prove various statements regarding factors and factorability.
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Guidance
In this lesson we will use induction to prove other kinds of statements. If you are not familiar with the concept of
mathematical induction, it is strongly recommend that you first complete the lesson: "Inductive Proofs".
There are two properties of integers and their factors that will be useful for the proofs in this lesson.
Property 1: If a is a factor of b, and a is a factor of c, then a is a factor of the sum b + c.
For example, the set of numbers a = 3, b = 6, and c = 9 satisfies the hypothesis because 3 is a factor of 6 and 3 is a
factor of 9. Property 1 states that therefore 3 is a factor of 9 + 6 = 15, which we know is true because 3 × 5
= 15. We can prove that this property is true for all integers if we think about what the term factor means. If
a is a factor of b, then there exists some integer M such that aM = b. Similarly if a is a factor of c, then there
exists some integer N such that aN = c. So we can write the sum b + c as aM + aN. We know
TABLE 1.37:
b + c = aM + aN = a(M +
N).
Because we can write the sum as a product of a and another number, a is a factor of the sum b + c.
Property 2: If a is a factor of b and b is a factor of c, then a is a factor of c.
We can prove this property in a similar manner. If a is a factor of b, then there exists some integer M such that aM
= b. If b is a factor of c, then there exists some integer N such that bN = c. We can write c in the following
manner:
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Chapter 1. Sequences, Series, and Mathematical Induction
c = bN = (aM)N = a(MN)
Therefore a is a factor of c because we can write c as the product of a and another integer, MN.
As noted above, these properties of integers are useful for proving statements about integers and factors via induction.
Example A
Prove that 3 is a factor of 4n - 1 for all positive integers n.
Solution
Proof by induction:
1. The base case: if n = 1, then 4n - 1 = 4 - 1 = 3. 3 is a factor of itself because 3 × 1 = 3 .
If the base case does not convince you, you can always test out additional values of n. For example, if n = 2, 42 1 = 16 - 1 = 15 = 5 × 3.
2. The inductive hypothesis: assume that 3 is a factor of 4k - 1.
3. The inductive step: show that 3 is a factor of 4k+1 - 1.
If 3 is a factor of 4k - 1, then there exists some integer M such that 3M = 4k - 1. We can write 4k+1 - 1 in a manner
that allows us to use the inductive hypothesis:
TABLE 1.38:
4k+1 - 1
4(4k -1) + 3
4(3M) + 3
Factor out 4, but add 3 in order to
keep the value of 4k+1 -1
Substitute: 3M = 4k - 1
Note that the substitution is the same as using property 2 above: if 3 is a factor of 4k - 1 , then 3 is a factor of 4(4k
- 1). Using the substitution simply makes the fact a bit more obvious.
This last step proves that 3 is a factor of 4k+1 - 1 , by property 1 above.
The technique of rewriting the k + 1 term can also be used to prove statements about polynomials and factors.
Example B
Prove that x - y is a factor of xn - yn for all positive integers n.
Note: Since we are talking about polynomials that are factorable now, not integers, then we say that if x - y is a
factor of xn - yn , then there exists a polynomial P such that P(x - y) = xn - yn .
Solution
Proof by induction:
1. The base case: If n = 1, we have xn - yn = x - y , and x - y is a factor of itself, as x - y = 1(x - y).
As we did above, we can also check n = 2 in order to convince ourselves. If n = 2, we have x2 - y2 = (x - y)(x + y),
so x - y is clearly a factor.
2. The inductive hypothesis: assume that x - y is a factor of xk - yk .
3. Show that x - y is a factor of xk+1 - yk+1 .
From the inductive step, we know that there is some polynomial P such that P(x - y) = xk - yk . We can rewrite xk+1
- yk+1 in a manner that allows use to use the inductive hypothesis:
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1.7. Induction and Factors
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TABLE 1.39:
xk+1 - yk+1
=xk+1 - xyk + xyk - yk+1
=x(xk - yk ) + yk (x - y)
=x(P(x - y)) + yk (x - y)
=Px(x - y) + y’k−1 (x - y)
Again, by property 1 above, this shows that x - y is a factor of xk+1 - yk+1 . Therefore we have shown that x - y is a
factor of xn - yn for all positive integers n.
Example C
a) Without adding, determine if 7 a factor of 49 + 70.
b) Consider the sum 23 + 54 = 77. Is 7 a factor of 77? What does this tell you about the first factor property in the
lesson?
Solution
a) Use Property 1 from the lesson: If a is a factor of b, and a is a factor of c, then a is a factor of the sum b + c.
7 is a factor of 49, since 7 × 7 = 49
7 is a factor of 70, since 7 × 10 = 70
Therefore 7 is a factor of 119, since 49 + 70 = 119
b) This is a test of the converse of Property 1, which would be "If a number is a factor of the sum, then it is a factor
of the factors of the sum"
7 is a factor of the sum: 77
7 is not a factor of 23 or 54
This tells us that the converse of the property is not necessarily true.
Vocabulary
A factor is a number or an expression that is multiplied with other factors to create a product.
A factorial refers to the product of the positive integers from 1 to some value n: n! = 1 × 2 × 3 × 4...× (n-1) × n
An inequality is a statement that two quantities are not equal.
A postulate is a statement that is accepted as true without proof.
Guided Practice
Questions
1) Prove that 9n - 1 is divisible by 8 for all positive integers n.
2) Prove that xn - 1 is divisible by x - 1 for all positive integers n.
3) Prove that n2 - n is even for all positive integers n.
4) Prove that 52n−1 + 1 is divisible by 6 for all positive integers n.
Solutions
1) Prove that 9n - 1 is divisible by 8 for all positive integers n.
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Chapter 1. Sequences, Series, and Mathematical Induction
TABLE 1.40:
If n = 1, 9n - 1 = 9 - 1 = 8 = 8(1)
Assume that 9k - 1 is divisible by 8.
Show that 9k+1 - 1 is divisible by 8.
1. Base case:
2. Inductive hypothesis:
3. Inductive step:
9k - 1 divisible by 8 ⇒ 8 W = (9k - 1) for some integer W
9k+1 - 1 = 9 (9k - 1) + 8 = 9 (8W) + 8, which is divisible by 8
2) Prove that xn - 1 is divisible by x - 1 for all positive integers n.
TABLE 1.41:
If n = 1, xk - 1 = x - 1 = (x - 1)(1)
Assume that xk - 1 is divisible by x - 1
Show that xk+1 - 1 is divisible by x - 1.
1. Base case:
2. Inductive hypothesis:
3. Inductive step:
xk - 1 divisible by x - 1 ⇒ P(x - 1) = (x k - 1) for some polynomial P
xk+1 - 1 = x(xk - 1) + (x - 1) = Px(x - 1) + (x - 1),which is divisible by x - 1
3) Prove that n2 - n is even for all positive integers n
TABLE 1.42:
1. Base case:
2. Inductive hypothesis:
3. Inductive step:
If n = 1, 12 - 1 = 1 - 1 = 0 = 2 × 0
Assume that k2 - k is even
Show that (k + 1)2 - (k + 1) is even.
If k2 - k is even, then k2 - k = 2 M for some integer M
(k + 1)2 - (k + 1) = k2 + 2k + 1 - k - 1 = k2 - k + 2k = 2M + 2k = 2(M + k) which is even because M + K is an integer.
4) Prove that 52n−1 + 1 is divisible by 6 for all positive integers n.
TABLE 1.43:
1. Base case:
2. Inductive hypothesis:
3. Inductive step:
If n = 1, 51 + 1 = 5 + 1 = 6 = 6(1)
Assume that 52k−1 + 1 is divisible by 6.
Show that 52(k+1)−1 + 1 is divisible by 6.
If 52k−1 + 1 is divisible by 6, then 52k−1 + 1 = 6M for some integer M.
52(k+1)−1 + 1 = 52k+1 + 1 = 52 (52k−1 + 1) - 24 = 52 (6M) - 24 which is divisible by 6.
37
1.7. Induction and Factors
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Practice
1. Without adding, determine if 7 a factor of 49 + 70
2. Consider the sum 23 + 54 = 77 Is 7 a factor of 77? What does this tell you about the first factor property in
the lesson?
3. Prove that any positive integer n >1 a) is prime or, b) can be represented as a product of prime factors.
4. Found within set "J" are all positive integers, from the number 1 to 2n. Prove that there are two numbers, one
that is a factor of another, from any (n = 1) numbers chosen from set "J"
5. Prove that (xn + x1n ) is also an integer for any positive integer n if the following is an integer: (x + 1x )
6. Prove the formula nk+m = nk−1 um + nk nm+1 for the sequence of Fibonacci numbers: n1 = 1, n2 = 1, uk+1 =
nk + nk−1 , k = 2, 3...
Prove the following identities
7. 12 + 22 + 32 + ... + n2 =
8.
9.
10.
11.
12.
n(n+1)(2n+1)
6
n2 (n+1)2
4
13 + 23 + 33 + ... + n3 =
1 · 2 · 3 + 2 · 3 · 4 + ... + n(n + 1)(n + 2) = n(n+1)(n+2)(n+3)
4
1x · 1! +2x · 2! +... + n · n! = (n + 1)! −1
2
= 13 + 23 + 33 + ... + n3
n2 (n+1)
4
(2n+1)
n(n + 1) 6 = 12 + 22 + 32 + ... + n2
Prove the following divisibilities:
is a factor of 1 + 2 + 3 + ... + n for all positive integers n
13. Prove that n (n+1)
2
14. Prove that 3 is a factor of n3 + 2n for all positive integers n.
2
15. Prove that n2 (n+1)
is a factor of 13 + 23 + 33 + ... + n3
4
38
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Chapter 1. Sequences, Series, and Mathematical Induction
1.8 Induction and Inequalities
Here you will learn about applying the concepts of mathematical induction to inequalities.
This is the third in a series of lessons on mathematical proofs. In this lesson we continue to focus mainly on proof
by induction, this time of inequalities, and other kinds of proofs such as proof by geometry.
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Guidance
The transitive property of inequality
Below, we will prove several statements about inequalities that rely on the transitive property of inequality:
If a <b and b <c , then a <c.
Note that we could also make such a statement by turning around the relationships (i.e., using “greater than”
statements) or by making inclusive statements, such as a ≥ b.
It is also important to note that this property of integers is a postulate, or a statement that we assume to be true. This
means that unlike the factor properties we used above, we need not prove the transitive property of inequality.
You encountered other useful properties of inequalities in earlier algebra courses:
Addition property: if a >b , then a + c >b + c.
Multiplication property: if a >b, and c >0 then ac >bc.
Example A
Prove that n! ≥ 2n for n ≥ 4
Solution
1. The base case is n = 4: 4! = 24, 24 = 16. 24 ≥ 16 so the base case is true.
2. Assume that k! ≥ 2k for some value of k such that k ≥ 4
3. Show that (k+1)! ≥ 2k+1
39
1.8. Induction and Inequalities
www.ck12.org
TABLE 1.44:
(k+1)! = k!(k+1)
≥ 2k (k +1)
≥ 2k (2)
Rewrite (k +1)! in terms of k !
Use step 2 and the multiplication property.
k +1 ≥ 5 >2, so we can use the multiplication property
again.
= 2k+1
Therefore n! ≥ 2n for n ≥ 4.
Example B
For what values of x is the inequality x >x2 true?
Solution
The inequality is true if x is a number between -1 and 1 but not 0.
Example C
Prove that 9n - 1 is divisible by 8 for all positive integers n.
Solution
TABLE 1.45:
1. Base case:
2. Inductive hypothesis:
3. Inductive step:
If n = 1, 9n - 1 = 9-1 = 8 = 8(1)
Assume that 9k - 1 is divisible by 8.
Show that 9k+1 - 1 is divisible by 8.
TABLE 1.46:
9k - 1 divisible by 8 ⇒ 8 W = (9k -1) for some integer W
9k+1 - 1 = 9(9k - 1) + 8 = 9(8W) + 8,which is divisible by 8
Vocabulary
A factor is a number or an expression that is multiplied with other factors to create a product.
Factorial refers to the product of the positive integers from 1 to some value n:
n! = 1 × 2 × 3 × 4...× (n-1) × n
An inequality is a statement that two quantities are not equal.
A postulate is a statement that is accepted as true without proof.
40
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Chapter 1. Sequences, Series, and Mathematical Induction
Guided Practice
1) Prove that 2n < n! for all positive integers n where n ≥ 4
2) Prove that n2 < 3n for all integers n > 2
3) Prove that 2n + 1 <2n for all integers n >3
Solutions
1) Use the three steps of proof by induction
Step 1) Base Case: 24 < 4!
2·2·2·2 < 1·2·3·4
16 < 24 ..... This checks out
Step 2) Assumption: 2k < k!
Step 3) Induction Step: starting with 2k < k! prove 2k (k + 1) < k! (k + 1)
2k (k + 1) < (k + 1)!
2 < k + 1 ..... If k ≥ 4 then this is true
2k · 2 < 2k (k + 1) ... Multiply both sides by 2k
2k+1 < 2k (k + 1)
2k+1 < (k + 1)!
∴ 2n < n! for all positive integers n where n ≥ 4
2) Use the three steps of proof by induction
Step 1) Base Case: (n = 1) 12 < 31 or, if you prefer, (n = 2) 22 < 32
Step 2) Assumption: k2 < 3k
Step 3) Induction Step: starting with k2 < 3k prove (k + 1)2 < 3k+1
k2 · 3 < 3k · 3
2k < k2 and 1 < k2 ..... assuming 2 < k as specified in the question
2k + 1 < 2k2 ..... combine the two statements above
k2 + 2k + 1 < 3k2 ..... add k2 to both sides
(k + 1)2 < 3k2
(k + 1)2 < 3 · 3k ..... from above
(k + 1)2 < 3k+1
∴ n2 < 3n for all integers n > 2
3) Prove that 2n + 1 <2n for all integers n >3
TABLE 1.47:
1. Base case:
2. Inductive hypothesis:
3. Inductive step:
If n = 3, 2(3) + 1 = 7, 23 = 8 : 7 <8, so the base case is
true.
Assume that 2k + 1 <2k for k >3
Show that 2(k + 1) + 1 <2k+1
41
1.8. Induction and Inequalities
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2(k + 1) + 1 = 2k + 2 + 1 = (2k + 1) + 2 <2k + 2 <2k + 2k = 2 (2k ) = 2k+1
Practice
1.
2.
3.
4.
5.
6.
5k < (k + 5)!
1k < (k + 1)!
4k < (k4)!
2k < (k + 2)!
For what values of x is the inequality x >x2 true?
Prove that 3n >n2 for all positive integers n.
Prove the following inequalities
1
1
1
1
7. n+1
+ n+2
+ n+3
+ ... + 2n
>
n
2
8. 2 ≥ n for n = 4, 5, 6, ...
9. 112 + 212 + 312 + ... + n12
13
24 (n
> 1)
1
n)
10. Given:x1 , ..., xn are positive numbers, prove the following: (x1 +...+x
≥ (x1 · ... · xn ) n
n
n
11. n! ≥ 3 for n = 7, 8, 9, ....
Geometric Induction
12. Prove that side length of a quadrilateral is less than the sum of all its other side lengths.
13. Prove that side length of a pentagon is less than the sum of all its other side lengths.
14. Prove that a square may be divided into any number of smaller squares greater than 5, without any remaining
squares.
15. Prove that it is possible to color all regions of a plane divided by several lines with two different colors, so that
any two neighbor regions contain a different color.
42
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Chapter 1. Sequences, Series, and Mathematical Induction
1.9 Sums of Finite Geometric Series
Here you will learn how to calculate the sum of a series when the difference between the members is not constant.
Anna is on a progressive workout plan, every day she adds 5% to her exercise time. If she starts by exercising 15mins
on the first day, how many minutes will she have exercised all together on day 45?
This is a geometric series, since the difference between the exercise time on any two days is greater than the
difference between any prior two days. You could just add: 15 + (15 · 1.05) + [(15 · 1.05) · 1.05]... and so on up
to 45, but that would be horribly tedious. In this lesson, you will learn how to answer a question like this will little
effort.
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Guidance
A finite geometric series is simply a geometric series with a specific number of terms. For example, consider the
series: 50 + 25 + 12.5 + ....The series is geometric: the first term is 50, and the common ration is (1/2).
The sum of the first two terms is 50 + 25 = 75. We can write this as S2 = 75
The sum of the first three is 50 + 25 + 12.5 = 87.5. We can write this as S3 = 87.5
To find the value of Sn in general, we could simply add together the first n terms in a series. However, this would
obviously be tedious for a large value of n. Given the regular pattern in a geometric series - every term is (1/r) of the
previous term, and the nth term is an = a1 rn−1 , we can use induction to prove a formula for Sn .
The sum of the first n terms in a geometric series is Sn =
a1 (1−rn )
1−r
For example, for the series 50 + 25 + 12.5 + ... , the sum of the first 6 terms is:
1
2
n)
50(1−( 21 )6 )
50(1− 64
) 50( 63
)
63
7
Sn = a1 (1−r
=
=
= 164 = 50 64
1
1
1−r
1 = 98 16
1−
2
2
2
The figure below shows the same calculation on a TI-83/4 calculator:
43
1.9. Sums of Finite Geometric Series
www.ck12.org
We can use this formula as long as the series in question is geometric.
Example A
Find the sum of the first 10 terms of a geometric series with a1 = 3 and r = 5.
Solution
The sum is 58,593.
n
7
)
)
3(1−78,125)
Sn = a1 (1−r
= 3(1−5
= 3(−78,124)
= 58, 593
1−r
1−5 =
−4
−4
Notice that because the common ratio in this series is 5, the terms get larger and larger. This means that for
increasing values of n the sums will also get larger and larger. In contrast, in the series with common ratio
(1/2), the terms gets smaller and smaller. This situation implies something important about the sum.
Example B
Find the sum of each series:
a) The first term of a geometric series is 4, and the common ratio is 3. Find S8 .
b) The first term of a geometric series is 80, and the common ratio is (1/4). Find S7 .
Solution
a) S8 =
b) S7 =
4(1−38 )
= 13, 120
1−3
1 7
80 1−( 4 )
1− 41
≈ 106.66
Example C
Prove the formula Sn =
a1 (1−rn )
1−r
by induction
Solution
1. If n = 1, the nth sum is the first sum, or a1 . Using the hypothesized equation, we get S1 =
This establishes the base case.
k)
2. Assume that the sum of the first k terms in a geometric series is Sk = a1 (1−r
1−r .
3. Show that the sum of the first k+1 terms in a geometric series is Sk+1 =
a1 (1−r1 )
1−r
=
a1 (1−r)
1−r
= a1 .
a1 (1−rk+1 )
.
1−r
TABLE 1.48:
Sk+1 = Sk + ak+1
=
44
a1 (1−rk )
1−r
+ a1 rk+1−1
The k + 1 sum is the kth sum, plus the k + 1 term
Substitute from step 2, and substitute the k + 1 term
www.ck12.org
Chapter 1. Sequences, Series, and Mathematical Induction
TABLE 1.48: (continued)
The k + 1 sum is the kth sum, plus the k + 1 term
Sk+1 = Sk + ak+1
=
=
=
=
=
a1
(1−rk )
rk (1−r)
a1
1−r +
1−r
k
a1 (1−r )+a1 rk (1−r)
1−r
a1 [1−rk +rk (1−r)]
1−r
a1 [1−rk +rk −rk+1 ]
1−r
a1 [1−rk+1 ]
1−r
The common denominator is 1 − r
Simplify the fraction
It is proven.
Therefore we have shown that Sn =
a geometric series.
a1 (1−rn )
1−r
for a geometric series. Now we can use this equation to find any sum of
TABLE 1.49:
Concept question wrap-up "Anna is on a progressive workout plan, every day she adds 5% to her exercise time.
If she starts by exercising 15mins on the first day, how many minutes will she have exercised all together on day
45?"
n)
Use the formula: Sn = a1 (1−r
1−r
Sn =
15(1−1.0545 )
1−1.05
Sn = 2395.5
minutes.
Vocabulary
A finite series has a defined ending value.
An infinte series does not have a defined ending value.
A geometric series is a series where the difference between terms increases or decreases between each pair of terms.
Guided Practice
Questions
1) Find the sum: 5 + 10 + 20 + ... + 640 (Hint: if an = 640 , what is n?)
2) Use a geometric series to answer the question:
In January, a company’s sales totaled $11,000. It is predicted that the company’s sales will increase 5% each month
for the next year. At this rate, what will be the total sales for the year?
3) Write the first 5 terms of the sequence: −5 · 43
n
N
4) Write the 3rd, 4th, and 6th terms of: (3)( 2 )
n−1
5) Find the sum of the series: ∑6n=1 − 32
Solutions
1) S8 =
5(1−28 )
1−2
= 1275
45
1.9. Sums of Finite Geometric Series
2) S12 =
11000(1−1.0512 )
1−1.05
www.ck12.org
= $175, 088.39
3) Just do the multiplication for each term n = 0 → n = 4
0
−5 · 43 to − 5 · 1 → −5 ..... for n = 0
1
−5 · 34 to − 5 · 34 → − 15
4 → −3.75 ..... for n = 1
2
9
→ − 45
−5 · 43 to − 5 · 16
16 → −2.8 ..... for n = 2
3
135
−5 · 34 to − 5 · 27
64 → − 64 → −2.1 ..... for n = 3
4
81
→ − 405
−5 · 34 to − 5 · 256
256 → −1.6 ..... for n = 4
∴ the first 5 terms are: −5, −3.75, −2.8, −2.1, −1.6
4) As with problem 3, just perform the operations on the indicated values of n:
√
√
3
3 2 → 33 → 27 → 5.2 ..... for n = 3
4
3 2 → 9 ..... for n = 4
6
3 2 → 33 → 27 ..... for n = 6
∴ the 3rd, 4th, and 6th terms are: 5.2, 9, 27
5) To find the sum of the series ∑6n=1 − 32
n−1
We could calculate all of the values for n = 1 → 6 and add them, getting:
9
−27
81
−243
1 + −3
2 + 4 + 8 + 16 + 32 =
k
Or we can use the formula: 1−r
1−r
6
1−( −3
2 )
= −133
−3
32
1−( 2 )
−133
32
Practice
Find the sum of the finite series. You may simply calculate the individual terms and add them, or you may use the
n)
formula: Sn = a1 (1−r
1−r
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
46
1
1 + − 21 + 14 + ... + 64
−6 + 12 − 24 + ... − 6144
(−4) + (−12) + (−36) + ... + (−2916)
9 + (−45) + 225 + ... + (−17, 578, 125)
∑5n=1 −(2)n−1
1 n−1
∑10
n=1 6 2
∑6n=1 8 · 3n−1
(−5) + 10 + (−20)
+ ... + (−1280)
9
9
9
(−9) + 2 + − 4 + ... + − 16
n−1
∑9n=1 4 · 12
2 n−1
∑10
n=1 4 · − 3
3
(−3) + − 23 + − 34 + ... + − 1024
(n−1)
∑11
n=1 −9 · (−2)
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Chapter 1. Sequences, Series, and Mathematical Induction
n−1
14. ∑9n=1 −9 · − 53
n−1
15. ∑6n=1 −2 · − 45
1 n−1
16. ∑11
n=1 8 · − 3
47
1.10. Sums of Infinite Geometric Series
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1.10 Sums of Infinite Geometric Series
Here you will explore geometric series with an infinite number of terms, and you will learn how to calculate the sum
of the terms in the series.
Sayber’s mom told him to clean his room on Saturday morning.
"But, MOM! It’s gonna take forever!" said Sayber.
"Oh, don’t be overly dramatic," said mom.
"I am NOT being dramatic!" Sayber said.
"If I start right now, it is going to take me at LEAST an hour to clean this half alone, then it will take another half
hour to clean half of the remainder, and 15 mins to clean half of THAT remainder... since I will always have
half left, I will never be done!"
Do you agree with Sayber? Will Sayber be stuck with a vacuum in his hand forever?
Tune in next week...
Watch This
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- James Sousa: Infinite Geometric Series
Guidance
Let’s return to the situation in the introduction: Poor Sayber is stuck cleaning his room. He cleans half of the room
in 60mins. Then he cleans half of what is left, 30 more minutes, half again for 15 more. If he keeps cleaning half of
the remaining area, how will he ever finish the room?
We know that the pieces have to add up to some finite time period (no matter what it feels like, Sayber CAN get the
room clean), but how is it possible for the sum of an infinite number of terms to be a finite number?
To find the sum of an infinite number of terms, we should consider some partial sums. Three partial sums, relatively
early in the series, could be: S2 = 90, S3 = 105, and S6 = 118.125 or 118 18
Now let’s look at larger values of n :
48
www.ck12.org
Chapter 1. Sequences, Series, and Mathematical Induction
TABLE 1.50:
S7
=
S8
=
S10
=
60(1−( 12 )7 )
≈ 119.06 minutes
1− 12
1 8
60(1−( 2 ) )
≈ 119.5 minutes
1− 12
60(1−( 12 )10 )
≈ 119.9 minutes
1− 21
As n approaches infinity, the value of Sn seems to approach 120 minutes. In terms of the actual sums, what is
happening is this: as n increases, the nth term gets smaller and smaller, and so the nth term contributes less and less
to the value of Sn . We say that the series converges, and we can write this with a limit:
TABLE 1.51:
60(1−( 21 )n )
1− 21
60(1−( 21 )n )
= limn→∞
limn→∞ Sn
= limn→∞
1
2
= limn→∞ 120 1 −
1 n
2
n
As n approaches infinity, the value of 12 gets smaller and
That is, the value of this expression approaches
smaller.
1 n
1 n
0. Therefore the value of 1 − 2 approaches 1, and 120 1 − 2
approaches 120(1) = 120.
Therefore, no matter how long the process continues, Sayber will not spend more than 2hrs cleaning the room. Of
course, it may SEEM like a lot more!
We can do the same analysis for the general case of a geometric series, as long as the terms are getting smaller and
smaller. This means that the common ratio must be a number between -1 and 1: |r| <1.
TABLE 1.52:
= limn→∞
limn→∞ Sn
=
a1
1−r ,
as
a1 (1−rn )
1−r
(1 − rn ) →
1
Therefore, we can find the sum of an infinite geometric series using the formula S =
a1
1−r .
When an infinite sum has a finite value, we say the sum converges. Otherwise, the sum diverges. A sum converges
only when the terms get closer to 0 after each step, but that alone is not a sufficient criterion for convergence. For
1
1
1
1
example, the sum ∑∞
n=1 n = 1 + 2 + 3 + 4 + .... does not converge.
Example A
Find the sum of the convergent series: 40 + −20 + 10 + −5 + ...
Solution
The common ratio is
−1
2 .
Therefore the sum converges to:
49
1.10. Sums of Infinite Geometric Series
www.ck12.org
TABLE 1.53:
40
1−(
−1
2
)
=
40
3
2
= 40
2
3
=
80
3
Example B
Determine if the series converges. If it converges, find the sum.
TABLE 1.54:
1
a) 1 + 13 + 19 + 27
+ ...
b) 3 + −6 + 12 + −24 + ...
Solution:
1
+ ... converges.
a) 1 + 13 + 19 + 27
The common ratio is (1/3) . Therefore the sum converges to:
TABLE 1.55:
1
1− 13
=
1
2
3
=
3
2
b) The series 3 + -6 + 12 + -24 + ... does not converge, as the common ratio is -2.
Remember that the idea of an infinite sum was introduced in the context of a realistic situation, albeit a paradoxical
one. We can in fact use infinite geometric series to model other realistic situations. Here we will look at another
example: the total vertical distance traveled by a bouncing ball.
Example C
A bouncing ball
A ball is dropped from a height of 20 feet. Each time it bounces, it reaches 50% of its previous height. What is the
total vertical distance the ball travels?
Solution
We can think of the total distance as the distance the ball travels down + the distance the ball travels back up. The
downward bounces form a geometric series:
20 + 10 + 5 +...
The upward bounces form the same series, except the first term is 10.
1 n−1
1 n−1
So the total distance is: ∑∞
+ ∑∞
.
n=1 20 2
n=1 10 2
Each sum converges, as the common ratio is (1/2). Therefore the total distance is:
50
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Chapter 1. Sequences, Series, and Mathematical Induction
TABLE 1.56:
10
+ 1−
1 =
2
40 + 20 = 60
20
1− 12
20
1
2
+
10
1
2
=
So the ball travels a total vertical distance of 60 feet.
Vocabulary
A geometric series is the sum of terms from a geometric sequence, which is a sequence of terms with a different
number or value between each pair of terms.
If the limit of the partial sums of series does not exist or is infinite, then the series diverges.
If the limit of the partial sums of series exists and is finite, the series converges.
Guided Practice
Questions
For questions 1 – 3, determine if the series converges or diverges. If it converges, find the sum.
1) -3 + 6 + -12 + 24 + ...
2) 240 + 60 + 15 + ...
3) 9 + 6 + 4 + (8/3) + ...
4) In this lesson, we proved the formula for the sum of a geometric series, Sn =
a1 (1−rn )
1−r
using induction.
Prove this formula without induction:
a) Let Sn = a1 + a1 r + a1 r2 + ... + a1 rn−1
b) Multiply Sn by r to obtain a second equation
c) Subtract the equations and solve for Sn .
5) A ball is dropped from a height of 40 feet, and each time it bounces, it reaches 25% of its previous height.
a) Find the total vertical distance the ball travels, using the method used in the lesson.
b) Find the total vertical distance the ball travels using a single series.
(Hint: write out several terms for each bounce. For example, the first bounce is: 40 feet down + 10 feet up = 50 feet
traveled.)
6) Below are two infinite series that are not geometric. Use a graphing calculator to examine partial sums. Does
either series converge?
a) 1 + 21 + 13 + 14 + ...
1
b) 1 + 14 + 19 + 16
+ ...
Solutions
1) The sum does not converge because r = -2.
2) The sum converges. S = 320.
51
1.10. Sums of Infinite Geometric Series
www.ck12.org
3) The sum converges. S = 27.
4) Follow the steps below:
(1)Sn = a1 + a1 r + a1 r2 + ... + a1 rn−1
(2) rSn = a1 r + a1 r2 + a1 r3 + ... + a1 rn
(3) Sn − rSn = a1 − a1 rn
⇒ Sn (1 − r) = a(1 − rn )
n)
⇒ Sn = a(1−r
(1−r)
5) a) ∑∞
n=1 40
b) ∑∞
n=1 50
1 n−1
1 n−1
+ ∑∞
n=1 20 4
4
1 n−1
4
= 66 32
= 66 23
6) a) This series does not converge.
b) This series converges around 1.65. (The actual sum is
π2
6)
Practice
1 n
1. Find the sum of the first 10 terms of ∑∞
n=1 5 using a graphing calculator.
1 n
2. Find the sum of the first 20 terms of ∑∞
using a graphing calculator.
n=1 5
3. Conjecture on the possible convergence of the series in questions 1 and 2.
Evaluate the infinite sum of the following geometric series:
4.
5.
6.
7.
8.
−2 + 1 − 12 + ...
96
−6 + 24
5 − 25 + ...
3
3
3 + 2 − 4 + ...
−6 + 4 − 83 + ...
1 + 12 + 14 + ...
Evaluate the infinite sum of the following geometric series:
10.
11.
12.
13.
14.
15.
52
1 (n−1)
∑∞
n=1 −3( 2 )
4 (n−1)
∑∞
n=1 −2( 7 )
∞
−4 (n−1)
∑n=1 7( 5 )
−1 (n−1)
∑∞
n=1 −9( 5 )
−5 (n−1)
∑∞
n=1 5( 7 )
1 (n−1)
∑∞
n=1 6( 5 )
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Chapter 1. Sequences, Series, and Mathematical Induction
1.11 Factorials and Combinations
Here you will review factorials and combinations in preparation for their use in Pascal’s Triangle and the Binomial
Theorem.
Kelly and Kyle are playing a card game, and Kyle is wondering why there never seem to be repeated hands. He
figures that since there are only 52 cards in the deck, and each hand has five cards, there really should be more
duplicate hands. After all, 5 is nearly 1/10 of 52.
Kelly tells him that she thinks there are thousands of possible combinations, and that she would be really surprised
to see the same 5 card hand twice or more in a given game.
Who is correct? Why?
Watch This
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- Khan Academy: Permutations and Combinations
Guidance
Recall that a factorial of a positive integer n is the product of n, and all of the positive integers less than n. We write
this as n! = n(n - 1)(n - 2) .. (3) (2) (1).
In order to develop the binomial theorem, we need to look at a related idea: combinations. If you have studied
probability, you may be familiar with combinations and permutations. A combination is the number of ways you
can choose r objects from a group of n objects, if the order of choosing does not matter. An example will help clarify
the idea of a combination:
Example A
In a class of 20 students, 3 students are going to be chosen to form a committee to plan a fieldtrip. How many
possible committees are there?
Solution
To answer this question, we need to figure out how many ways we can choose groups of 3 students from the 20 on
the class. The order of choosing does not matter. That is, if I choose Amy, Juan, and Nina, it is the same as
choosing Juan, then Amy, then Nina, or any other ordering of the three students.
In general, we can find the number of combinations of r objects chosen from n objects by the following:
53
1.11. Factorials and Combinations
www.ck12.org
TABLE 1.57:
n
r
=
(Note that there are two different symbols for combinations: nCr and
is used on the TI-83/84.
n
r
nCr
=
n!
r!(n−r)!
You can use either one, though nCr is what
Therefore the number of combinations of 3 people from 20 people in the class is
TABLE 1.58:
20!
3!17!
= 1140
Example B
How many different groups of 3 cards can be chosen from 20 different cards, assuming order does not matter? Use
a graphing calculator.
Solution
20
3
To find
Press: 20 <TI font_MATH>and then move right to the PRB menu.
Press 3. This takes you back to the main screen. You should see 20 nCr .
Now press 3 >,TI font_ENTER>.
You should see the answer, 1140.
Example C
Calculate by hand: How many different 4-person teams can be made from 7 people?
Solution
Smaller numbers, such as these, are not too difficult to calculate by hand.
7
4
=
7!
4!3!
=
7·6·5·4·3·2·1
4·3·2·1·3·2·1
=
7·6·5
3·2·1
= 7 · 5 = 35.
Canceling factors in the numerator and denominator simplifies the calculation.
TABLE 1.59:
54
www.ck12.org
Chapter 1. Sequences, Series, and Mathematical Induction
TABLE 1.59: (continued)
Concept question wrap-up Recall in the introduction we learned of the disagreement between Kelly and Kyle?
Kelly thinks there are thousands of possible 5-card combinations in a deck of cards, Kyle thinks there should not
be all that many.
This is a classic combinations problem in the form "52, choose 5":
n!
Using the formula: nCr = nr = r!(n−r)!
52!
We get: 52C5 = 52
5 = 5!(52−5)!
52!
5!(47)!
Simplify more: 52·51·50·49·48
5·4·3·2·1
And once more: 311,875,200
120
Simplify to:
Gives: 2, 598, 960
Looks like Kelly undershot by quite a bit too!
Vocabulary
A combination refers to choosing one or more objects from a set when the order of choosing does not matter.
A factorial refers to the product of the positive integers from 1 to some value n: 1 × 2 × 3 × 4 × .... × (n - 1) × n
= n!
Guided Practice
Questions
1) Simplify:
(m+3)!
m+1 !
2) Demonstrate that:
2(4)C2
= 2(4C2 ) + 42
3) In how many ways can i pick 6 jelly beans from a container containing 10 jelly beans?
4) At the carnival, you decide to play a game of chance. You buy 15 tickets for the game. You have a 75% chance of
winning each time you play the game. What is the probability that you will win exactly 8 of the 15 games?
5) Explain why the following equality holds.
(n+1)Cr
=n Cr +n Cr−1
Solutions
1) First we need to expand the numerator and denominator:
(m+3)(m+2)(m+1)(m)(m−1)...(1)
(m+1)(m)(m−1)...(1)
(m + 3)(m + 2) ..... Cancel common factors
m2 + 5m + 6 ..... Simplify
2) Evaluate both sides:
= 28
4C2 = 6
42 = 16
8C2
55
1.11. Factorials and Combinations
www.ck12.org
Check: 28 = 2(6) + 16
28 = 12 + 16
28 + 28 - the equality holds.
3) We calculate using the formula:10C6 =
10!
6!(10−6)!
10! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)
10! = 3628800
6! = (1)(2)(3)(4)(5)(6)
6! = 720
(10 − 6)! = (1)(2)(3)(4)
(10 − 6)! = 24
∴10 C6 =
3628800
720·24
→ 210
4) The chance is about 4%
For each of the 15 games, there is 75% chance
of winning, and a 25% chance of losing.
15
8
The probability of exactly 8 wins is 8 (.75) (.25)7 ≈ 0.039 → 3.9%
Note that this is only the probability of winning exactly 8 games, no more, no less.
5) Proof of equality:
a) remove one item from the set.
b) either the (r) item we want to come out of the remaining (n)items, or
c) we choose (r - 1) item from the (n) remaining items and we include the one item we removed.
Practice
Simplify and Evaluate the Factorials.
1.
2.
(b−2)!
(b−5)!
(a+2)!
(a+1)!
7!
3!3!
9!
8!
3.
4.
5. 4! +3!
6. 6!
5!
Show that the equality holds by evaluating both sides of the equation:
7. 4C2 =4 C4−2
8. (7+1)C4 =7 C4 +7 C(4−1)
Explain conceptually why the following equality holds:
9.
2nC2
Solve
10. 8C5
56
= 2(nC2 ) + n2
www.ck12.org
Chapter 1. Sequences, Series, and Mathematical Induction
11. 6C3
12. In a class of 200 students, 25 will be chosen randomly to participate in a research study. How many possible
groups of 25 students can be chosen? (Hint: use a calculator!)
13. A die is rolled 10 times. What is the probability of rolling exactly four 4’s? (Hint: the probability of rolling a
4 is 1/6. The probability of not rolling a 4 is 5/6.)
14. The local TV station forecasts a 30% chance of rain every day for the next week. What is the probability that
it will rain on exactly 6 out of the next 7 days?
15. Consider the following situation: a basketball player is going to attempt to make 20 free throws. She is
assuming that she has an 80% chance of making each shot. What is the probability that she will make exactly
19 out of 20 shots?
16. Two students were
this problem. Student A says that the problem is a Bernoulli trial, and that the
discussing
19 (.2)1 Student B disagrees, and says that the situation is not a Bernoulli trial. What
probability is 20
(.8)
19
reasoning might student B use to support his argument?
57
1.12. Binomial Theorem and Expansions
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1.12 Binomial Theorem and Expansions
Here you will learn about Pascal’s Triangle, and you will explore the connection between the triangle and the
Binomial Theorem used to expand binomials.
Multiplying binomials is not a terribly difficult exercise, but it can certainly be time consuming with higher exponents, for example:
Calculate: (x − 3)2 is pretty easy:
(x − 3) · (x − 3)
(x2 − 6x + 9)
...not bad
BUT...
Calculate: (x − 3)5
This is: (x − 3) · (x − 3) · (x − 3) · (x − 3) · (x − 3)
First we "foil" the first two terms to get (x2 − 6x + 9)
2nd, we multiply (x2 − 6x + 9) by (x − 3), yielding: (x3 − 9x2 + 27x − 27)
3rd, we multiply (x3 − 9x2 + 27x − 27) by (x − 3)
... and so on.
Definitely do-able, but a nightmare of a job, particularly by hand.
Isn’t there an easier way?
Watch This
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- James Sousa: Binomial Expansion using Pascal’s Triangle
Guidance
There is a particular pattern in combinations that is seen in the expansion of polynomials of the form (x + y)n .
This pattern is most commonly displayed in a triangle:
58
www.ck12.org
Chapter 1. Sequences, Series, and Mathematical Induction
This triangle is referred to as Pascal’s triangle, named after mathematician Blaise Pascal, although other mathematicians before him worked with these numbers. The numbers in the triangle can be used to generate more rows: notice
that if you add two consecutive numbers, you get the number between and below them in the next row.
n
We can generalize this pattern as follows: r−1
+ nr = n+1
r .
Binomial Expansion
To expand a binomial is to multiply all of the factors. The resulting polynomial is in standard form. For example:
(x + y)2 = (x + y) (x + y) = x2 + xy + xy + y2 = x2 + 2xy + y2
If we expand (x + y)3 , we get:
TABLE 1.60:
(x + y)3
= (x + y) (x + y) (x + y)
= (x + y) (x2 + 2xy + y2 )
= x3 + 2x2 y + xy2 + x2 y + 2xy2 + y3
= x3 + 3x2 y + 3xy2 + y3
Notice that the coefficients of each polynomial correspond to a row of Pascal’s triangle.
Also notice that the exponents of x descend, and the exponents of y ascend with each term. These are key aspects of
the Binomial Theorem.
The Binomial Theorem
The binomial theorem can be stated using a summation:
TABLE 1.61:
(x + y)n = ∑nr=0
n
r
n−r r x y
This is a very succinct way of summarizing the pattern in a binomial expansion. Let’s return to (x + y)3 to see how
the theorem works.
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TABLE 1.62:
3 3−0 0
y + 31 x3−1 y1 +
0 x
3 3−2 2
y + 33 x3−3 y3
2 x
= 1x3 · 1 + 3x2 y + 3xy2 + 1 · x0 · y3
(x + y)3
=
= x3 + 3x2 y + 3xy2 + y3
Again, the exponents on x descend from 3 to 0. The exponents on y ascend from 0 to 3. The coefficients on the terms
correspond to row 3 of Pascal’s triangle. These coefficients are, not surprisingly, referred to as binomial coefficients!
Given this theorem, we can expand any binomial without having to multiply all of the factors.
Finding a specific term in a Binomial Expansion
Finding a term in an expansion can be used to answer a particular kind of probability question.
Consider an experiment, in which there are two possible outcomes, such as flipping a coin. If we flip a coin over
and over again, this is referred to as a Bernoulli trial. In each flip (“experiment”), the probability of getting a head
is 0.5, and the probability of getting a tail is 0.5. (Note: this is true for flipping a coin, but not for other situations.
That is, it’s not always “50-50 chance!) Now say we flip a coin 25 times. What is the probability of getting exactly
10 heads?
The answer to this question is a term of a binomial expansion. That is, the probability of getting 10 heads from 25
10
15
coin tosses is: 25
10 (0.5) (0.5) ≈ 0.0974, or about a 10% chance.
Example A
Use the binomial theorem to expand each polynomial:
TABLE 1.63:
a.(2x + a)4
b. (x - 3)5
Solution
a.(2x + a)4
TABLE 1.64:
= 1(2x)4 (a)0 + 4(2x)3 (a)1 + 6(2x)2 (a)2 + 1(2x)0 (a)4
= 16x4 + 32x3 a + 24x2 a2 + 8x’a3 + a4
Note that it is easier to simply use the numbers from the appropriate row of the triangle than to write out all of the
coefficients as combinations. However, if n is large, it may be easier to use the combinations.
b. (x - 3)5
TABLE 1.65:
= 1(x)5 (-3)0 + 5(x)4 (-3)1 + 10(x)3 (-3)2 + 10(x)2 (-3)3 +
5(x)1 (-3)4 + 1(x)0 (-3)5
= x5 - 1x4 a + 90x3 - 270x2 + 405x - 243
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Chapter 1. Sequences, Series, and Mathematical Induction
Notice that in this expansion, the terms alternate signs. This is the case because the second term in the binomial
is -3. When expanding this kind of polynomial, be careful with your negatives!
We can also use the Binomial Theorem to identify a particular term or coefficient.
Example B
Identify the 3rd term of the expansion of (2x + 3)6 .
Solution
The 3rd term is
6
2
(2x)4 32 = 15 · 16x4 · 9 = 2160x4
Keep in mind that row 6 of Pascal’s triangle starts with 60 , so the coefficient of the third term in the expansion is
6
6
0
4
2
2 . Also keep in mind that the first term includes (2 x) and 3 , so the third term includes (2x) and 3 .
Example C
What is the coefficient of x2 in the expansion of (3x + 3)3
Solution
Use the binomial theorem for the second term:
2
3−2
3C2 (3x) (3)
2
1
3(3x) (3) = 81
Vocabulary
The Binomial Theorem is an efficient formula for calculating the expansion of binomials.
Binomial Expansion is the process of raising a binomial such as (x + 2) to a power. The process can be time
consuming, particularly with higher exponents.
Pascal’s Triangle is a pyramid of sorts constructed with the coefficients of binomials as they are expanded. It is a
convenient reference.
Guided Practice
1) Use Pascal’s triangle to find the coefficient of x2 y in (x − y)3
2) Expand: (x + 3)3
3) Expand: (3x − 3)6
4) Find the coefficient ofx3 in (x − y)3
Solutions
1) First, create Pascal’s triangle. Remember to add 1’s on the ends and the sum of the two numbers above to get the
new numbers.
1
1−1
1−2−1
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1−3−3−1
We only had to go up to the third row, we are looking for the 2nd term, so we count over 1 + 1 from the left to get
our coefficient of 3. Don’t forget to start with the 1st term.
We use the formula for the second term:
3C2 x
2 (−y)3−2
3x2 (−y)1
So we get an answer of: -3
2) Start by expanding out each term separately:
Use the formula for term 0:
3C3 x
3 (3)3−3
= x3 (3)0 = x3
...and term 1:
3C2 x
2 (3)3−2
= 3x2 (3)1 = 9x2
...and term 2:
3C1 x
1 (3)3−1
= 3x1 (3)2 = 27x
...and term 3:
3C0 x
0 (3)3−0
= x0 (3)3 = 27
So we end up with:
x3 + 9x2 + 27x + 27
3) We will expand out each term separately:
We use the formula for term 0:
6
6−6
6C6 (3x) (−3)
= (3x)6 (−3)0 = 729x6
...and term 1:
5
6−5
6C5 (3x) (−3)
= 6(3x)5 (−3)1 = −4374x5
...and term 2:
4
6−4
6C4 (3x) (−3)
= 15(3x)4 (−3)2 = 10, 935x4
...and term 3:
3
6−3
6C3 (3x) (−3)
= 20(3x)3 (−3)3 = −14, 580x3
...and term 4:
2
6−2
6C2 (3x) (−3)
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= 15(3x)2 (−3)4 = 10, 935x2
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Chapter 1. Sequences, Series, and Mathematical Induction
...and term 5:
6C1 (3x)1 (−3)6−1 = 6(3x)1 (−3)5 = −4374x
...and term 6:
0
6−0
6C0 (3x) (−3)
= (3x)0 (−3)6 = 729
The expanded polynomial is:
729x6 − 4374x5 + 10, 935x4 − 14, 580x3 + 10, 935x2 − 4374x + 729
4) We use the formula for the third term:
3
3−3
3C3 x (−y)
3
0
x (−y)
Giving us our answer: 1
Practice
Questions
1.
2.
3.
4.
5.
6.
7.
Expand:
Expand:
Expand:
Expand:
Expand:
Expand:
Expand:
(x + 3a)4
(y + 21 )5
(2x − a)6
(x + y)6
(3x + 1)5
(x + y)5
(2x + 2)4
Find the Term
8. Find the 3rd term in the expansion (3x + 2a)9 .
9. Find the 7th term in the expansion of (4x − 12 a)10 .
Use Pascal’s Triangle to find the coefficient:
10. What is the coefficient of x3 y2 in the expansion of (x + y)5 ?
11. What is the coefficient of x4 in the expansion(3x + 1)4 ?
12. What is the coefficient of x in the expansion:(2x + 2)5 ?
What is the coefficient of the expansions?
13.
14.
15.
16.
17.
What is the coefficient of x2 in the expansion of (x + 1)6 ?
What is the coefficient of x5 in the expansion of (2x + 1)5 ?
What is the coefficient of x in the expansion of (3x + 2)3 ?
What is the coefficient of x6 in the expansion of (x + y)6 ?
What is the coefficient of x in the expansion of (2x + 1)6 ?
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Summary
This chapter covers the application and creation of arithmetic and geometric formulas. Students are introduced to
Sigma (sum) notation and are taught to calculate the sum of a partial or infinite series.
Later in the chapter, students will learn about induction and inductive proofs. The chapter wraps up with combinations/permutations and factorials and an introduction to the binomial theorem and binomial expansion.
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