Download Chapter 11: Series and Patterns

Chapter Eleven: Series and Patterns
Section One: Sequences and Series
A sequence is an ordered list of numbers. Each number in the list is called a term. We use three dots
(ellipsis) to show that a sequence continues with no end. These are called infinite sequences. A finite
sequence has a last term. We name each term using subscripts ( t1, t2 , t3 , t4 ,..., tn1, tn ).
We can use a rule to describe an infinite sequence using the set of natural numbers as the domain.
When we define the sequence in this way, we call the rule an explicit formula.
EX1: Write the first six terms of the sequence defined by the explicit formula tn  4n  5
A sequence can also be defined using a recursive formula. These formulas define each term in relation
to the term before it. For example, let t1  2 and tn  tn1  3 . This rule says that the first term is 2 and
each term is 3 more than the term before it. This recursive formula defines that sequence
2,5,8,11,14,17... .
EX2: Write the first six terms of the sequence defined by the recursive formula t1  4 and tn  3tn1  5 ,
where n  2 .
A series is the summation of the terms of a sequence. We use summation notation, which uses the
greek letter sigma  , to represent a series.
4
 2n means that we will plug in the numbers from 1 to 4
n 1
into the function 2n. The sigma tells us to then sum the terms.
EX3: Evaluate the following series
6
a.
 3k
k 1
b.
1 6
k
2 k 1
The following properties apply to summations
1. Scalar Product:
3n  3  n

2.

Addition:  m  n    m   n
EX4: Find the following summations two different ways
3
a.
 2k
k 1
k
3
b.
k 1
2
 k
The following formulas give us ways to sum up three common types of functions
n
Constant Series:
c  nc
k 1
n(n  1)
2
k 1
n
n  n  1 2n  1
Quadratic Series:  k 2 
6
k 1
n
Linear Series:
k 
EX5: Evaluate the following using the three formulas
5
a.
4
k 1
4
b.
x
2
x 1
4
c.
 3n
n 1
2
 2n  4
Section Two: Arithmetic Sequences
An arithmetic sequence is a sequence whose successive terms differ by the same number, called the
common difference. The recursive formula for an arithmetic sequence is tn  tn1  d , where d is the
common difference. All arithmetic sequences are linear! The formula for an arithmetic sequence is given
by tn  t1  (n  1)d where d is the common difference.
EX1: Find the explicit formula defined by t1  7 and tn  tn1  6 . Then find the 10th term of the
sequence.
We can also find the explicit formula using the slope-intercept form y  mx  b . All you need to find
the equation is two points.
EX2: Find the explicit formula defined by t1  3 and tn  tn1  4 . Then find the 7th term of the sequence.
EX3: A new car costs $16,000. The value of the car depreciates $1500 per year. Use a recursive formula
to find the value of the car after 5 years. Use the explicit formula for the nth term of an arithmetic
sequence to find the value of the car after 7 years.
EX4: Find the 15th term of the arithmetic sequence in which t5  7 and t10  22 .
The missing terms between any two nonconsecutive terms in an arithmetic sequence are referred to as
the arithmetic means.
EX5: Find the five arithmetic means between 6 and 60.
Section Three: Arithmetic Series
An arithmetic series is the sum of an arithmetic sequence. When summing an arithmetic sequence, we
can use a coupling method. By coupling the first and last term, we can begin to see a pattern.
10
 2k  2  4  6  8  10  12  14  16  18  20
k 1
Because of the commutative property of addition, notice how we can couple the different numbers
together
10
 2k   2  20    4  18   6  16   8  14   10  12 
k 1
We now have 5 groups of numbers (half of n) that add up to 22 ( t1  t10 ). So we can simply find the sum
by finding 5  22  110 .
This method can lead to the following formula: S n 
EX1: Given 16, 12, 8, 4,…, find S11
21
EX2:
  5  4k 
k 1
n
 t1  tn 
2
Section Four: Geometric Sequence
A geometric sequence is a sequence in which the ratio of successive terms is the same number, called
the common ratio. Therefore, the recursive formula for an geometric sequence is tn  r  tn1 , where r is
the common ratio. Geometric sequences are exponential functions.
EX1: Write the first five terms of the geometric sequence if t1  4 and tn  2  tn1 .
Since geometric sequences are exponential functions we can define it with the following explicit
formula: tn  t1  r n1 .
EX2: Find the seventh term of the sequence defined by t1  5 and tn  3  tn1 .
EX3: A car that costs $26,500 depreciates, and its value at the end of a given year is 90% of its value at
the end of the preceding year. Use a recursive method to find the car’s value after five years. Use the
formula for the nth term of a geometric sequence to find the value of the car after 8 years.
EX4: Find the 10th term of the geometric sequence in which t2  48 and t5  384 .
The geometric means are the missing terms between any two nonconsecutive terms of a geometric
sequence.
EX5: Find three geometric means between 324 and 4.
Section Five: Geometric Series and Mathematical Induction
A geometric series is the sum of a geometric sequence. The sum of the first n terms of a geometric
series can be found with the following formulas:
 1 rn 
Sn  t1 

 1 r 
EX1: Given the series 400  300  225  168.75  ... , find S16 to the nearest tenth.
EX2: Given the series 6 14.4  34.56  82.944  ... , find S 8 to the nearest tenth.
EX3: To save for a newborn child’s college education, a family plans to invest $2500 at the beginning of
every year for the next 18 years. Suppose that the investment earns 6% annual interest, compounded
annually. How much will the investment be worth at the end of the 18th year?
7
EX4:
 4  5
k 1
k 1
To use mathematical induction to prove that a statement is true, you use the following steps:
1. Show that the statement is true for n  1
2. Assume the statement is true for k
3. Use this to show the statement is true for k+1
EX5: Prove the following statements by using mathematical induction
a. For every natural number n, 4  8  12  ...  4n  2n(n  1)
b. For every natural number n, 5  11  17  ...   6n 1  n(3n  2)
Section Six: Infinite Geometric Series
An infinite geometric series is a geometric series with an infinite number of terms. A partial sum is the sum of just
a portion of the infinite series. We can find the partial sum by using the formula from section 5,
 1 rn 
Sn  t1 
.
 1 r 
Sometimes the partial sums get closer and closer to a fixed value; they converge. When a sequence
diverges it does not close in on a single number. When the common ratio of the series is between -1 and
1 (not 0) then the sequence will converge. When a sequence converges, we can find the sum of the
entire infinite series by using the formula S 
t1
.
1 r
We will now take a look at a series that converges. The following series 32,16,8,4,2,1,0.5,0.25,...
1
. Its partial sums are 32,48,56,60,62,63,63.5,63.75,...
2
has a ratio of
Notice, that the graph of partial sums level off at 64. Therefore, we can say that the sum of the infinite
sequence is 64. (There is an asymptote line there.)
EX1: Find the sum of the infinite series if it exists.
a. 8,9.6,11.52,13.824,...
b.
4,2, 1,0.5,...
c.
2

1
k 1

d.
1
5
k 1
k 1
k 1
Section Seven: Pascal’s Triangle
In this section we will explore a famous set of patterns known as Pascal’s triangle.
1
1
1
2
1
3
1
1
1
4
7
3
10
6
1
1
6
5
1
1
15
4
10
20
21
1
35
5
1
15
6
35
21
1
7
1
The rows of the triangle correspond to our study of combinations.
0
1
2
3
4
5
6
7
C0
C0
C0
C1
C1
6
7
C2
6
7
C3
3
4
5
C4
4
5
6
C3
C3
C3
C3
7
C2
C2
C2
C2
C2
2
3
4
5
C1
C1
C1
C1
C1
1
2
3
4
5
6
7
C0
C0
C0
C0
C0
C4
C4
5
6
7
C4
C5
C5
C5
6
7
C6
C6
7
C7
EX1: Use Pascal’s Triangle to find the following combinations
a. 5 C2
b.
c.
C7
8 C4
7
EX2: Suppose that a fair coin is tossed 5 times. In how many ways can exactly 0, 1, 2, 3, 4, and 5 heads
appear?
In review, if we flip a single coin there is 2 outcomes. If we flip 2 coins there are 2  2  4 different
outcomes. If we flip 3 coins there are 2  2  2  8 different outcomes. If we flip n coins there are 2 n
different outcomes. We can use these facts and the combinations from the triangle to find the
probabilities of different coin combinations.
EX3: Find the probability of getting exactly 2 heads or exactly 3 heads when a fair coin is tossed 5 times.
Give your answer to the nearest hundredth.
EX4: Find the probability of getting more than 4 heads when a fair coin is tossed 7 times. Give your
answer to the nearest hundredth.
Section Eight: The Binomial Theorem
In previous lessons we reviewed how to square binomials such as ( x  3) 2 using FOIL. This is a pretty
easy task but it’s not quite so easy to multiply out higher powered binomials.
( x  3) 4 
( x  3) 2 ( x  3) 2 
( x 2  6 x  9)( x 2  6 x  9) 
x 4  6 x 3  9 x 2  6 x 3  36 x 2  54 x  9 x 2  54 x  81 
x 4  12 x 3  54 x 2  108 x  81
We can use Pascal’s Triangle to make this task a little bit easier.
EX1: Use Pascal’s triangle to expand the following binomials
a. ( x  y)6
b.
c.
( x  2)5
(3x  2 y)7
EX2: Find the indicated term in the expansion of (2 x  4)9
a. the fourth term
b. the seventh term