Coburn J.W. Algebra and Trigonometry
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1020 CHAPTER 11 Additional Topics in Algebra 11-4
Solution
a. b. c.
Now try Exercises 39 through 44
EXAMPLE 4
Computing a Specified Term from a Sequence Defined Using Factorials
Find the third term of each sequence.
a. b.
Solution
a. b.
Now try Exercises 45 through 50
C. Series and Partial Sums
Sometimes the terms of a sequence are dictated by context rather than a formula. Con-
sider the stacking of large pipes in a storage yard. If there are 10 pipes in the bottom
row, then 9 pipes, then 8 (see Figure 11.1), how many pipes are in the stack if there is
a single pipe at the top? The sequence generated is 10, 9, 8, ..., 3, 2, 1 and to answer
the question we would have to compute the sum of all terms in the sequence.When the
terms of a finite sequence are added, the result is called a finite series.
Finite Series
Given the sequence a
1
, a
2
, a
3
, a
4
,..., a
n
, the sum of the terms is called a finite series
or partial sum and is denoted S
n
:
EXAMPLE 5
Computing a Partial Sum
Given find the value of
a. S
4
and b. S
7
.
Solution
Since we eventually need the sum of the first seven terms (for Part b), begin by
writing out these terms: 2, 4, 6, 8, 10, 12, and 14.
a.
b.
Now try Exercises 51 through 56
56
2 4 6 8 10 12 14
S
7
a
1
a
2
a
3
a
4
a
5
a
6
a
7
20
2 4 6 8
S
4
a
1
a
2
a
3
a
4
a
n
2n,
S
n
a
1
a
2
a
3
a
4
p
a
n
20
112152!
3!
11235
#
4
#
3!4
3!
6
8
3
4
c
3
112
3
32132 14!
3!
a
3
3!
2
3
c
n
112
n
12n 12!
n!
a
n
n!
2
n
1 495 72
6
6
990
2
9
#
8
6!
3!5!
6
#
5!
3!5!
11!
8!2!
11
#
10
#
9
#
8!
8!2!
9!
7!
9
#
8
#
7!
7!
WORTHY OF NOTE
Most calculators have a
factorial option or key. On the
TI-84 Plus it is located on a
submenu of the key:
PRB (option) 4: !
MATH
MATH
B. You’ve just learned how
to work with recursive
sequences and sequences
involving a factorial
Figure 11.1
C. You’ve just learned how
to find the partial sum of a
series
College Algebra & Trignometry—
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11-5 Section 11.1 Sequences and Series 1021
D. Summation Notation
When the general term of a sequence is known, the Greek letter sigma can be used
to write the related series as a formula. For instance, to indicate the sum of the first
four terms of we write This result is called summation or
sigma notation and the letter i is called the index of summation. The letters j, k, l,
and m are also used as index numbers, and the summation need not start at 1.
EXAMPLE 6
Computing a Partial Sum
Compute each sum:
a. b. c.
Solution
a.
b.
c.
Now try Exercises 57 through 68
If a definite pattern is noted in a given series expansion, this process can be
reversed, with the expanded form being expressed in summation notation using the nth
term.
EXAMPLE 7
Writing a Sum in Sigma Notation
Write each of the following sums in summation (sigma) notation.
a. b.
Solution
a. The series has five terms and each term is an odd number, or 1 less than a
multiple of 2. The general term is and the series is
b. This is an infinite sum whose terms are multiples of 3. The general term is
but the series starts at 2 and not 1. The series is
Now try Exercises 69 through 78
Since the commutative and associative laws hold for the addition of real numbers,
summations have the following properties:
q
j2
3j.a
n
3n,
5
n1
12n 12.a
n
2n 1,
6 9 12
p
1 3 5 7 9
9 16 25 36 18
6
k3
112
k
k
2
112
3
#
3
2
112
4
#
4
2
112
5
#
5
2
112
6
#
6
2
60
60
30
60
20
60
15
60
12
60
137
60
5
j1
1
j
1
1
1
2
1
3
1
4
1
5
5 8 11 14 38
4
i1
13i 22 13
#
1 22 13
#
2 22 13
#
3 22 13
#
4 22
6
k3
112
k
k
2
5
j1
1
j
4
i1
13i 22
4
i1
13i 22.a
n
3n 2,
©
WORTHY OF NOTE
By varying the function given
and/or where the sum
begins, more than one
acceptable form is possible.
For example 7(b)
also works.
q
n1
13 3n2
College Algebra & Trignometry—
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1022 CHAPTER 11 Additional Topics in Algebra 11-6
Properties of Summation
Given any real number c and natural number n,
(I)
If you add a constant c “n” times the result is cn.
(II)
A constant can be factored out of a sum.
(III)
A summation can be distributed to two (or more) sequences.
(IV)
A summation is cumulative and can be written as a sum of smaller parts.
The verification of property II depends solely on the distributive property.
Proof:
expand sum
factor out c
write series in summation form
The verification of properties III and IV simply uses the commutative and asso-
ciative properties. You are asked to prove property III in Exercise 91.
EXAMPLE 8
Computing a Sum Using Summation Properties
Recompute the sum from Example 6(a) using summation properties.
Solution
property III
property II
property I
result
Now try Exercises 79 through 82
E. Applications of Sequences
To solve applications of sequences, (1) identify where the sequence begins (the initial
term) and (2) write out the first few terms to help identify the nth term.
38
1 2 3 4 10; 31102 2142
3
4
i1
i
4
i1
2
4
i1
13i 22
4
i1
3i
4
i1
2
4
i1
13i 22
c
n
i1
a
i
c1a
1
a
2
a
3
p
a
n
2
n
i1
ca
i
ca
1
ca
2
ca
3
p
ca
n
m
i1
a
i
n
im1
a
i
n
i1
a
i
; 1 m 6 n
n
i1
1a
i
b
i
2
n
i1
a
i
n
i1
b
i
n
i1
ca
i
c
n
i1
a
i
n
i1
c cn
D. You’ve just learned how
to use summation notation to
write and evaluate series
College Algebra & Trignometry—
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11-7 Section 11.1 Sequences and Series 1023
EXAMPLE 9
Solving an Application — Accumulation of Stock
Hydra already owned 1420 shares of stock when her company began offering
employees the opportunity to purchase 175 discounted shares per year. If she made
no purchases other than these discounted shares each year, how many shares will
she have 9 yr later? If this continued for the 25 yr she will work for the company,
how many shares will she have at retirement?
Solution
To begin, it helps to simply write out the first few terms of the sequence. Since she
already had 1420 shares before the company made this offer, we let be
the inaugural element, showing (after 1 yr, she owns 1595 shares). The
first few terms are 1595, 1770, 1945, 2120, and so on. This supports a general term
of
After 9 years After 25 years
After 9 yr she would have 2995 shares. Upon retirement she would own 5795
shares of company stock.
Now try Exercises 85 through 90
5795 2995
a
25
1595 1751242 a
9
1595 175182
a
n
1595 1751n 12.
a
1
1595
a
0
1420
E. You’ve just learned to
use sequences to solve
applied problems
Studying Sequences and Series
TECHNOLOGY HIGHLIGHT
To support a study of sequences and series, we can use a graphing
calculator to generate the desired terms. This can be done either on the
home screen or directly into the LIST feature of the calculator. On the
TI-84 Plus this is accomplished using the “seq(” and “sum(”
commands, which are accessed using the keystrokes
(LIST) and the screen shown in Figure 11.2. The “seq(” feature is option
5 under the OPS submenu (press 5) and the “sum(” feature is
option 5 under the MATH submenu (press 5).
To generate the first four terms of the sequence and
to find the sum, the home screen and press
5 to place “seq(” on the home screen. This command requires
four inputs: a
n
(the nth term), variable used (the calculator can work
with any letter), initial term and the last term. For this example the
screen reads “seq ” with the result being the four terms
shown in Figure 11.3. To find the sum of these terms, we simply
precede the seq command by “sum(,” and two
methods are shown in Figure 11.4.
Each of the following sequences have some interesting properties
or mathematical connections. Use your graphing calculator to
generate the first 10 terms of each sequence and the sum of the first
10 terms. Next, generate the first 20 terms of each sequence and the
sum of these terms. What conclusion (if any) can you reach about the
sum of each sequence?
Exercise 1: Exercise 2: Exercise 3: a
n
1
12n 1212n 12
a
n
2
n1n 12
a
n
1
3
n
1x
2
1, x, 1, 42
1x
2
1, x, 1, 42,
STAT
2nd
CLEAR
a
n
n
2
1,
STAT
2nd
Figure 11.2
Figure 11.3
Figure 11.4
College Algebra & Trignometry—
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CONCEPTS AND VOCABULARY
1024 CHAPTER 11 Additional Topics in Algebra 11-8
Fill in the blank with the appropriate word or phrase.
Carefully reread the section if needed.
1. A sequence is a(n) of numbers listed in a
specific .
2. A series is the of the numbers from a
given sequence.
3. When each term of a sequence is larger than the
preceding term, the sequence is said to be .
4. When each term of a sequence is smaller than the
preceding term, the sequence is said to be .
5. Describe the characteristics of a recursive sequence
and give one example.
6. Describe the characteristics of an alternating
sequence and give one example.
DEVELOPING YOUR SKILLS
Find the first four terms, then find the 8th and 12th
term for each nth term given.
7. 8.
9. 10.
11. 12.
13. 14.
15. 16.
17. 18.
19. 20.
21. 22.
Find the indicated term for each sequence.
23. 24.
25. 26.
27. 28. a
n
3a
1
3
b
n1
; a
7
a
n
2a
1
2
b
n1
; a
7
a
n
112
n1
2n 1
; a
5
a
n
112
n1
n
; a
5
a
n
1n 22
2
; a
9
a
n
n
2
2; a
9
a
n
112
n
2
n
a
n
112
n
2
n
a
n
112
n1
2n
2
1
a
n
112
n
n1n 12
a
n
1
n
2
a
n
1
n
a
n
a
2
3
b
n
a
n
a
1
2
b
n
a
n
a1
1
n
b
n
a
n
n
n 1
a
n
112
n
n
a
n
112
n
n
a
n
2n
3
12a
n
3n
2
3
a
n
2n 3a
n
2n 1
29. 30.
31.
32.
Find the first five terms of each recursive sequence.
33.
34.
35.
36.
37.
38. e
c
1
1, c
2
2
c
n
c
n1
1c
n2
2
2
•
c
1
64, c
2
32
c
n
c
n2
c
n1
2
e
a
1
2
a
n
a
n1
16
e
a
1
1
a
n
1a
n1
2
2
3
e
a
1
3
a
n
2a
n1
3
e
a
1
2
a
n
5a
n1
3
a
n
1
12n 1212n 12
; a
5
a
n
1
n12n 12
; a
4
a
n
an
1
n
b
n
; a
9
a
n
a1
1
n
b
n
; a
10
11.1 EXERCISES
College Algebra & Trignometry—
cob19529_ch11_1017-1096.qxd 12/8/08 21:31 Page 1024
Simplify each factorial expression.
39. 40. 41.
42. 43. 44.
Write out the first four terms in each sequence.
45. 46.
47. 48.
49. 50.
Find the indicated partial sum for each sequence.
51. 52.
53. 54.
55. 56.
Expand and evaluate each series.
57. 58.
59. 60.
61. 62.
63. 64.
4
i2
i
2
4
i1
i
2
2
5
k1
112
k
2
k
7
k1
112
k
k
5
k1
1k
2
12
5
k1
12k
2
32
5
i1
12i 32
4
i1
13i 52
a
n
n
n 1
; S
4
a
n
1
n
; S
5
a
n
3n 1; S
6
a
n
2n 1; S
8
a
n
n
2
; S
7
a
n
n; S
5
a
n
2
n
n!
a
n
n
n
n!
a
n
1n 32!
12n2!
a
n
1n 12!
13n2!
a
n
n!
1n 32!
a
n
n!
1n 12!
10!
3!7!
8!
2!6!
6!
3!3!
9!
7!2!
12!
10!
8!
5!
65. 66.
67. 68.
Write each sum using sigma notation. Answers are not
necessarily unique.
69.
70.
71.
72.
For the given general term a
n
, write the indicated sum
using sigma notation.
73.
74.
75. third partial sum
76. sixth partial sum
77. sum for to 7
78. sum for to 6
Compute each sum by applying properties of summation.
79. 80.
81. 82.
4
k1
12k
3
52
4
k1
13k
2
k2
6
i1
13 2i2
5
i1
14i 52
n 2a
n
n
2
;
n 3a
n
n
2
n
;
a
n
2n 1;
a
n
n
2
3
;
a
n
n
2
1
n 1
; S
4
a
n
n 3; S
5
1 8 27 64 125 216
1 4 9 16 25 36
5 10 15 20 25
4 8 12 16 20
6
k2
112
k1
k
2
1
8
k3
112
k
k1k 22
7
j3
j
2
j
7
j3
2j
11-9 Section 11.1 Sequences and Series 1025
WORKING WITH FORMULAS
83. Sum of
The sum of the first n terms of the sequence
defined by 4, 7, 10,...,
. . . is given by the formula shown. Find
S
5
using the formula, then verify by direct
calculation.
13n 22,
a
n
3n 2 1,
a
n
3n 2: S
n
n(3n 1)
2
84. Sum of
The sum of the first n terms of the sequence defined
by 5, 8, 11,..., ... is
given by the formula shown. Find S
8
using the
formula, then verify by direct calculation. Observing
the results of Exercises 83 and 84, can you now state
the sum formula for a
n
3n 0?
13n 12,a
n
3n 1 2,
a
n
3n 1: S
n
n(3n 1)
2
College Algebra & Trignometry—
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1026 CHAPTER 11 Additional Topics in Algebra 11-10
EXTENDING THE CONCEPT
91. Verify that a summation may be distributed to two
(or more) sequences. That is, verify that the
following statement is true:
.
Surprisingly, some of the most celebrated numbers in
mathematics can be represented or approximated by a
series expansion. Use your calculator to find the partial
n
i1
1a
i
b
i
2
n
i1
a
i
n
i1
b
i
sums for and for the summations
given, and attempt to name the number the summation
approximates:
92. 93.
94.
n
k1
1
2
k
n
k1
1
3
k
n
k0
1
k!
n 12n 4, n 8,
MAINTAINING YOUR SKILLS
95. (6.7) Solve
96. (2.5) Set up the difference quotient for
then rationalize the numerator.
f 1x2 1x
,
csc x sin
a
2
xb1
97. (7.2) Given a triangle where
find the three
corresponding angles.
98. (6.3) Solve the system using a matrix
equation. •
25x y 2z 14
2x y z 40
7x 3y z 13
c 0.5 m,b 0.3 m, and
a 0.4 m,
APPLICATIONS
Use the information given in each exercise to determine
the nth term for the sequence described. Then use the
nth term to list the specified number of terms.
85. Blue-book value: Steve’s car has a blue-book
value of $6000. Each year it loses 20% of its value
(its value each year is 80% of the year before). List
the value of Steve’s car for the next 5 yr. (Hint: For
we need the next five terms.)
86. Effects of inflation: Suppose inflation (an increase
in value) will average 4% for the next 5 yr. List the
growing cost (year by year) of a DVD that costs
$15 right now. (Hint: For we need the
next five terms.)
87. Wage increases: Latisha gets $5.20 an hour for
filling candy machines for Archtown Vending.
Each year she receives a $0.50 hourly raise. List
Latisha’s wage for the first 5 yr. How much will
she make in the fifth year if she works 8 hr per day
for 240 working days?
88. Average birth weight: The average birth weight of
a certain animal species is 900 g, with the baby
gaining 125 g each day for the first 10 days. List
the infant’s weight for the first 10 days. How much
does the infant weigh on the 10th day?
a
1
15,
a
1
6000,
a
n
89. Stocking a lake: A local fishery stocks a large
lake with 1500 bass and then adds an additional
100 mature bass per month until the lake nears
maximum capacity. If the bass population
grows at a rate of 5% per month through
natural reproduction, the number of bass in the
pond after n months is given by the recursive
sequence
How many bass will be in the lake after
6 months?
90. Species preservation: The Interior Department
introduces 50 wolves (male and female) into a
large wildlife area in an effort to preserve the
species. Each year about 12 additional adult
wolves are added from capture and relocation
programs. If the wolf population grows at a rate
of 10% per year through natural reproduction,
the number of wolves in the area after n
years is given by the recursive sequence
How many
wolves are in the wildlife area after 6 years?
w
0
50, w
n
1.10w
n1
12.
b
0
1500, b
n
1.05b
n1
100.
College Algebra & Trignometry—
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11.2 Arithmetic Sequences
Similar to the way polynomials fall into certain groups or families (linear, quadratic,
cubic, etc.), sequences and series with common characteristics are likewise grouped.
In this section, we focus on sequences where each successive term is generated by
adding a constant value, as in the sequence 1, 8, 15, 22, 29, ..., where 7 is added to
a given term in order to produce the next term.
A. Identifying an Arithmetic Sequence and Finding
the Common Difference
An arithmetic sequence is one where each successive term is found by adding a fixed
constant to the preceding term. For instance 3, 7, 11, 15, . . . is an arithmetic sequence,
since adding 4 to any given term produces the next term. This also means if you take
the difference of any two consecutive terms, the result will be 4 and in fact, 4 is called
the common difference d for this sequence. Using the notation developed earlier, we
can write where a
k
represents any term of the sequence and rep-
resents the term that follows a
k
.
Arithmetic Sequences
Given a sequence a
1
, a
2
, a
3
,..., a
k
, ..., a
n
, where k, and
if there exists a common difference d such that
then the sequence is an arithmetic sequence.
The difference of successive terms can be rewritten as (for
to highlight that each following term is found by adding d to the previous term.
EXAMPLE 1
Identifying an Arithmetic Sequence
Determine if the given sequence is arithmetic.
a. 2, 5, 8, 11,... b.
Solution
a. Begin by looking for a common difference Checking each pair
of consecutive terms we have
This is an arithmetic sequence with common difference
b. Checking each pair of consecutive terms yields
Since the difference is not constant, this is not an arithmetic sequence.
Now try Exercises 7 through 18
EXAMPLE 2
Writing the First k Terms of an Arithmetic Sequence
Write the first five terms of the arithmetic sequence, given the first term a
1
and the
common difference d.
a. and b. and d
1
3
a
1
1
2
d 4a
1
12
3
6
1
2
2
6
1
3
4
3
5
6
8
6
5
6
5
6
1
2
5
6
3
6
d 3.
5 2 3
8 5 3
11 8 3
and so on.
d a
k1
a
k
.
1
2
,
5
6
,
4
3
,
7
6
, p
k 12a
k1
a
k
d
a
k1
a
k
d,
k 6 n,
n a
k1
,
a
k1
d a
k1
a
k
,
Learning Objectives
In Section 11.2 you will learn how to:
A. Identify an arithmetic
sequence and its
common difference
B. Find the nth term of an
arithmetic sequence
C. Find the nth partial sum
of an arithmetic
sequence
D. Solve applications
involving arithmetic
sequences
11-11 1027
College Algebra & Trignometry—
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1028 CHAPTER 11 Additional Topics in Algebra 11-12
Solution
a. and Starting at add to each new term to
generate the sequence: 12, 8, 4, 0,
b. and Starting at and adding to each new term will
generate the sequence:
Now try Exercises 19 through 30
B. Finding the nth Term of an Arithmetic Sequence
If the values a
1
and d from an arithmetic sequence are known, we could generate the
terms of the sequence by adding multiples of d to the first term, instead of adding d to
each new term. For example, we can generate the sequence 3, 8, 13, 18, 23 by adding
multiples of 5 to the first term
current term initial coefficient of common
term difference
It’s helpful to note the coefficient of d is 1 less than the subscript of the current
term (as shown): This observation leads us to a formula for the nth term.
The nth Term of an Arithmetic Sequence
The nth term of an arithmetic sequence is given by
where d is the common difference.
EXAMPLE 3
Finding a Specified Term in an Arithmetic Sequence
Find the 24th term of the sequence 0.1, 0.4, 0.7, 1, ....
Solution
Instead of creating all terms up to the 24th, we determine the constant d and use
the nth term formula. By inspection we note and
nth term formula
substitute 0.1 for a
1
and 0.3 for d
eliminate parentheses
simplify
To find the 24th term we substitute 24 for n:
substitute 24 for n
result
Now try Exercises 31 through 42
EXAMPLE 4
Finding the Number of Terms in an Arithmetic Sequence
Find the number of terms in the arithmetic sequence 2, ..., 411.19,12,5,
7.0
a
24
0.31242 0.2
0.3n 0.2
0.1 0.3n 0.3
0.1 1n 120.3
a
n
a
1
1n 12d
d 0.3.a
1
0.1
a
n
a
1
1n 12d
5 1 4.
23 3 1425
a
5
a
1
4d
18 3 1325
a
4
a
1
3d
13 3 1225
a
3
a
1
2d
8 3 1125
a
2
a
1
1d
3 3 1025
a
1
a
1
0d
a
1
3:
1
2
,
5
6
,
7
6
,
3
2
,
11
6
1
3
a
1
1
2
d
1
3
.a
1
1
2
4
4a
1
12,d 4.a
1
12
A. You’ve just learned how
to identify an arithmetic
sequence and its common
difference
S
S
College Algebra & Trignometry—
cob19529_ch11_1017-1096.qxd 12/8/08 21:33 Page 1028
11-13 Section 11.2 Arithmetic Sequences 1029
Solution
By inspection we see that and As before,
nth term formula
substitute 2 for a
1
and for d
simplify
Although we don’t know the number of terms in the sequence, we do know
the last or nth term is Substituting for a
n
gives
substitute for a
n
solve for n
There are 60 terms in this sequence.
Now try Exercises 43 through 50
If the term a
1
is unknown but a term a
k
is given, the nth term can be written
since
(the subscript of the term a
k
and coefficient of d sum to n).
EXAMPLE 5
Finding the First Term of an Arithmetic Sequence
Given an arithmetic sequence where and find the common
difference d and the value of a
1
.
Solution
At first it seems that not enough information is given, but recall we can express a
13
as the sum of any earlier term and the appropriate multiple of d. Since a
6
is known,
we write (note as required).
a
1
is unknown
substitute 0.9 for a
13
and 0.55 for a
6
subtract 0.55
solve for d
Having found d, we can now solve for a
1
.
nth term formula for
substitute 0.9 for a
13
and 0.05 for d
simplify
solve for a
1
The first term is and the common difference is
Now try Exercises 51 through 56
C. Finding the nth Partial Sum of an Arithmetic Sequence
Using sequences and series to solve applications often requires computing the sum of
a given number of terms. Consider the sequence a
1
, a
2
, a
3
, a
4
,..., a
n
with common
difference d. Use S
n
to represent the sum of the first n terms and write the original
series, then the series in reverse order underneath. Since one row increases at the same
rate the other decreases, the sum of each column remains constant, and for simplic-
ity’s sake we choose to represent this sum.
S
n
a
1
a
2
a
3
p
a
n2
a
n1
a
n
S
n
a
n
a
n1
a
n2
p
a
3
a
2
a
1
2S
n
1a
1
a
n
2 1a
1
a
n
2 1a
1
a
n
2
p
1a
1
a
n
2 1a
1
a
n
2 1a
1
a
n
2
a
1
a
n
d 0.05.a
1
0.3
a
1
0.3
0.9 a
1
0.6
0.9 a
1
1210.052
n 13 a
13
a
1
12d
d 0.05
0.35 7d
0.9 0.55 7d
a
13
a
6
7d
13 6 7a
13
a
6
7d
a
13
0.9,a
6
0.55
n k 1n k2a
n
a
k
1n k2d
60 n
411 411 7n 9
411411.
7n 9
7 2 1n 12172
a
n
a
1
1n 12d
d 7.a
1
2
B. You’ve just learned how
to find the nth term of an
arithmetic sequence
add
columns
vertically
↓
College Algebra & Trignometry—
cob19529_ch11_1017-1096.qxd 12/8/08 21:33 Page 1029