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7.1 Arithmetic Sequences
Sequence: An ordered list of numbers identified by a pattern or rule that may stop at
some number or continue indefinitely. Ex. 1, 2, 4, 8, ... Ex. 3, 7, 11, 15
Term (of a sequence): A single value or object in a sequence. Each number in a sequence is
called a term . Each term is identified by its position in the sequence
An arithmetic sequence is a sequence in which the difference between consecutive terms is
a constant. The difference between consecutive terms of an arithmetic sequence is called a
common difference . Ex. 13, 11, 9, 7, ...
In this example the common difference is -2
The following sequence is arithmetic, ie. the terms increase by the same amount.
Figure 1
Figure 2
Recursive Formula:
Figure 3
or
Figure 4
...
Figure n
General Term of an Arithmetic Sequence:
A recursion formula is a formula by which
each term of a sequence is generated from
the preceding term or terms.
t1 = a , tn = tn-1 + d
tn = a + (n-1)d
,n > 1
A recursion formula has 2 parts:
First part: begins the sequence ex. t1 = 1
Here we can calculate any term (nth term) in a
sequence providing we know the first term ('a')
and the common difference ('d').
Second part: used to write the terms, one
after the other. ex. tn = tn-1 + 2, n > 1
A recursion formula allows you to calculate any
term provided you know the preceding term.
How many blocks would be needed for the 47th term?
If you had 135 blocks, what figure number could you make?
Ex1.
Determine if the sequence is arithmetic.
If it is, state the general term, and the recursive formula.
a) 35, 33, 31, 29, ...
b)
c)
Ex2. Determine the first 3 terms. If arithmetic, state the common
difference.
a) tn = 3n2 + 1
b) f(n) = 10 - n
Ex3. Determine the number of terms in the following arithmetic sequence.
0.6, 0.72, 0.84, ... , 7.44
.
7.1 Assignment: p. 424 #5-11, 13,15,16 (parts acde for all)
7.2 Geometric Sequences
Geometric Sequence: a sequence that has a common ratio between consecutive terms.
The common ratio "r" can be found by taking any term and dividing it by the term
before.
3, 12, 48, 192, ...
General Term of a Geometric Sequence:
Recursive Formula:
t1 = a , tn = rtn-1
Ex1.
,n > 1
tn = arn-1
A company has 3 kg, of radioactive material that must be stored. After one year,
95% of the radioactive material remains. How much radioactive material will be left
at the end of 10 years?
Ex2. Determine if the following sequences are arithmetic, geometric or neither. If they are
arithmetic or geometric state the general term for the sequence.
a) 31, 32, 34, 37, ...
b) 29, 19, 9, -1, ...
c) 128, 96, 72, 54, ...
Ex3. Given a geometric sequence with t7 = -192 and t9 = -768, determine t16
Ex4. Determine the general term and the recursive formula for the following
geometric sequence.
1, 0.2, 0.04, ...
7.2 Assignment: p. 430 #3, 5-12 (parts acde) 13, 14ab
7.3/7.4 Exploring Recursive Sequences and Creating Rules to Define Sequences
A recursion formula is a formula by which each term of a sequence is generated from the
preceding term or terms.
A recursion formula is a “recipe” for generating the terms, starting with the first term
A recursion formula allows you to calculate any term provided you know the preceding term
Examples:
1. Write the first 4 terms for the following sequences given their recursion formula.
a)
b)
c)
d)
The Fibonacci sequence is the sequence of numbers 1, 1, 2, 3, 5, 8, 13, 21, 34, ...
Each number after the first two numbers, 1 and 1, is the sum of the preceding two numbers
a. Write a recursion formula for the Fibonnaci sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, ...
b. Determine the 10th , 11th , and 12th , term.
Write a recursion formula for the following sequences.
a) 2, -6, 18, -54, ...
b) 4, 5, 7, 10, 14, ...
A sequence is an ordered list of numbers that may or may not follow a predictable pattern.
A sequence has a general term if an algebraic rule using the term number, n, can
be found to generate each term.
Write the first 3 terms in each sequence. Then determine the 10th term of each
sequence.
a)
b)
Determine the general term of the sequence.
a) 2, 6, 12, 20, 30 ....
b)
p. 439 #4a, 7, 11 p. 443 #3
7.5 Arithmetic Series
Arithmetic Series: The sum of the terms of an arithmetic sequence.
For the story of Carl Friedrich Gauss and how it relates to the sum of an arithmetic
series.
http://www.coolmath.com/algebra/19­sequences­series/06­gauss­problem­arithmetic­series­01
What Gauss actually did was find a pattern by reversing the series and
Adding vertically:
1 + 2 + 3 +...+ 98 + 99 + 100
+ 100 + 99 + 98 +...+ 3 + 2 + 1
101 + 101 + 101 +...+ 101 + 101 + 101
This represents twice the sum of the series that we want, therefore:
or
We can use this concept to develop a general formula for the sum of an arithmetic
series with first term "a" and a common difference "d".
The sum of a general arithmetic series:
S = a + (a+d) + (a+2d) + ... + [a+(n­3)d] + [a+(n­2)d] + [a+(n­1)d]
S = [a+(n­1)d] + [a+(n­2)d] + [a+(n­3)d] + ... + (a+2d) + (a+d) + a
2S = [2a + (n­1)d] + [2a + (n­1)d] + [2a + (n­1)d] +... +[2a + (n­1)d] + [2a + (n­1)d] + [2a + (n­1)d]
2S = n x [2a + (n­1)d]
Therefore the formula for the sum of a general arithmetic series is:
Ex.1: Determine the sum of the first 30 terms of 2 + 7 + 12 +...
Ex.2: Determine the sum of: 4 + 2.5 + 1 + ... ­ 33.5
Ex.3: Determine the sum of an arithmetic series in which t5 = 21 and the last term t36 = ­72
.
Ex.4: In an arithmetic series S1 = 1, S2 = 3, S3 = 6. Determine S12 and t100
Assignment 7.5: p. 452 #4­7, 10­12, 15,16 7.6 Geometric Series
Recall a Geometric Sequence: 2, 10, 50, 250, 1250, 6250, ... with a = 2, r = 5 and whose general term is: t n = ar n­1
Geometric Series is just the sum of a geometric sequence. Therefore the geometric series for the above geometric sequence is: 2 + 10 + 50 + 250 + 6250 + ... with a = 2, r = 5 and whose general term is: t n = ar n­1
What if we wanted the sum of the first 6 terms of the series above:
S = 2 + 10 + 50 + 250 + 1250 + 6250
5S = 10 + 50 + 250 + 1250 + 6250 + 31250
Trick: multiply through by "5"
Subtract:
S ­ 5S = 2 ­ 31250
Therefore: ­4S = 2 ­ 31250
S = ­31248 / ­4 S = 7812
Now let's find the Sum for a general geometric series: S
a
S­rS
=
a
Factor out the S: S(1­r) =
a
or
+
rS =
multiply through by "r"
Subtract:
=
S
=
­
a ­ arn
1 ­ r
ar
+
ar2 +
ar3 + ... + arn­2 +
arn­1
ar
+
ar2 +
ar3 + ... + arn­2 +
arn­1 +
arn
­
arn
arn
or
Sn =
a(1 ­ rn)
1 ­ r
or
Sn =
a(rn ­ 1)
r ­ 1
Ex.1: Calculate t8 and S8 for each of the following geometric series:
a)
­3 + 6 ­ 12 + ...
b) 6 + 8 + ...
.
c) 5 ­ 10x + 20x2 ­ ...
Ex.2: Find the sum of:
960 + 480 + 240 + ... + 15
Assignment 7.6: p. 459 #3­8,11 7.7 PASCAL'S TRIANGLE AND THE BINOMIAL THEOREM
Pascal's Triangle is a triangular arrangement of number with a 1 in the first
row, and 1 and 1 in the second row. Each number in the succeeding rows is
the sum of the two numbers above it in the preceding row
Position of Terms: A term in Pascal's Triangle can be represented by tn,r
where n is the horizontal row number and r is the diagonal row number. So we can represent the triangle as such
Each term is equal to the sum of the two terms immediately above it.
tn,r = tn-1,r-1 + tn-1,r
Ex. Express t5,3 + t5,4 as a single term
Binomial Expansion
Ex. Expand the following
* Compare the nth power to the nth row's elements in Pascal's Triangle. Now
compare these to the coefficients of the expanded polynomial. Look for
other patterns in the expanded polynomials.
* Powers of binomials can be expanded using patterns. The coefficients on
the expansion of (a + b)n can be found in row n of Pascal's Triangle.
Ex. Expand the following using the Binomial Theorem and Pascal's Triangle
a.
b.
c.
d.
e.
Assignment 7.7: p. 466 #4, 5, 10