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11.1 Sequences and Series

Find terms of sequences given the nth term.
Look for a pattern in a sequence and try to determine a general term.
Convert between sigma notation and other notation for a series.
Construct the terms of a recursively defined sequence.

In this section, we discuss sets or lists of numbers, considered in order, and their sums.

Sequences

Suppose that $1000 is invested at 4%, compounded annually. The amounts to which the account will grow after 1 year, 2 years, 3 years, 4 years, and so on, form the following sequence of numbers:

(1) ⏐ ↓ $1040.00 (2) ⏐ ↓ $1081.00 (3) ⏐ ↓ $1124.86 (4) ⏐ ↓ $1169.86.

$\begin{array}{c} (1) & (2) & (3) & (4) \\ ↓ & ↓ & ↓ & ↓ \\ $1040.00 & $1081.00 & $1124.86 & $1169.86. \end{array}$

We can think of this as a function that pairs 1 with $1040.00, 2 with $1081.60, 3 with $1124.86, and so on. A sequence is thus a function, where the domain is a set of consecutive positive integers beginning with 1.

If we continue to compute the amounts of money in the account forever, we obtain an infinite sequence with function values

$1040.00, $1081.60, $1124.86, $1169.86, $1216.65, $1265.32, \dots .

$$1040.00, $1081.60, $1124.86, $1169.86, $1216.65, $1265.32, \dots .$

The dots “…” at the end indicate that the sequence goes on without stopping. If we stop after a certain number of years, we obtain a finite sequence:

$1040.00, $1081.60, $1124.86, $1169.86.

$$1040.00, $1081.60, $1124.86, $1169.86.$

Consider the sequence given by the formula

a (n) = 2 n, or a n = 2 n .

$a (n) = 2^{n}, or a_{n} = 2^{n} .$

Some of the function values, also known as the terms of the sequence, follow:

a 1 a 2 a 3 a 4 a 5 = = = = = 21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32.

$\begin{array}{rcl} a_{1} & = & 2^{1} = 2, \\ a_{2} & = & 2^{2} = 4, \\ a_{3} & = & 2^{3} = 8, \\ a_{4} & = & 2^{4} = 16, \\ a_{5} & = & 2^{5} = 32. \end{array}$

The first term of the sequence is denoted as a1 $a_{1}$ , the fifth term as a5 $a_{5}$ , and the nth term, or general term, as an $a_{n}$ . This sequence can also be denoted as

2, 4, 8, \dots, or as 2, 4, 8, \dots, 2 n, \dots .

$2, 4, 8, \dots, or as 2, 4, 8, \dots, 2^{n}, \dots .$

Example 1

Find the first 4 terms and the 23rd term of the sequence whose general term is given by an=(−1)nn2 $a_{n} = {(- 1)}^{n} n^{2}$ .

Solution

We have an=(−1)nn2 $a_{n} = {(- 1)}^{n} n^{2}$ , so

a 1 a 2 a 3 a 4 a 23 = = = = = (- 1) 1 \cdot 12 = - 1, (- 1) 2 \cdot 22 = 4, (- 1) 3 \cdot 32 = - 9, (- 1) 4 \cdot 42 = 16, (- 1) 23 \cdot 232 = - 529.

$\begin{array}{rcl} a_{1} & = & {(- 1)}^{1} \cdot 1^{2} = - 1, \\ a_{2} & = & {(- 1)}^{2} \cdot 2^{2} = 4, \\ a_{3} & = & {(- 1)}^{3} \cdot 3^{2} = - 9, \\ a_{4} & = & {(- 1)}^{4} \cdot 4^{2} = 16, \\ a_{23} & = & {(- 1)}^{23} \cdot 23^{2} = - 529. \end{array}$

Now Try Exercise 1.

Note in Example 1 that the power (−1)n ${(- 1)}^{n}$ causes the signs of the terms to alternate between positive and negative, depending on whether n is even or odd. This kind of sequence is called an alternating sequence.

We can graph a sequence just as we graph other functions. Consider the function given by f(x)=x+1 $f (x) = x + 1$ and the sequence whose general term is given by an=n+1 $a_{n} = n + 1$ . The graph of f(x)=x+1 $f (x) = x + 1$ is shown on the left below. Since the domain of a sequence is a set of positive integers, the graph of a sequence is a set of points that are not connected. Thus if we use only positive integers for inputs of f(x)=x+1 $f (x) = x + 1$ , we have the graph of the sequence an=n+1 $a_{n} = n + 1$ , as shown on the right below.

Finding the General Term

When only the first few terms of a sequence are known, we do not know for sure what the general term is, but we might be able to make a prediction by looking for a pattern.

Example 2

For each of the following sequences, predict the general term.

1,2–√,3–√, 2,… $1, \sqrt{2}, \sqrt{3}, 2, \dots$
−1, 3, −9, 27, −81,…
2, 4, 8,…

Solution

These are square roots of consecutive integers, so the general term might be n−−√ $\sqrt{n}$ .
These are powers of 3 with alternating signs, so the general term might be (−1)n3n−1 ${(- 1)}^{n} 3^{n - 1}$ .
If we see the pattern of powers of 2, we will see 16 as the next term and guess 2n $2^{n}$ for the general term. Then the sequence could be written with more terms as

$2, 4, 8, 16, 32, 64, 128, \dots .$ $2, 4, 8, 16, 32, 64, 128, \dots .$

If we see that we can get the second term by adding 2, the third term by adding 4, and the next term by adding 6, and so on, we will see 14 as the next term. A general term for the sequence is n2−n+2 $n^{2} - n + 2$ , and the sequence can be written with more terms as

$2, 4, 8, 14, 22, 32, 44, 58, \dots .$ $2, 4, 8, 14, 22, 32, 44, 58, \dots .$

Now Try Exercise 19.

Example 2(c) illustrates that, in fact, you can never be certain about the general term when only a few terms are given. The fewer the number of given terms, the greater the uncertainty.

Sums and Series

Example 3

For the sequence −2, 4, −6, 8, −10, 12, −14,…, find each of the following.

S1 $S_{1}$
S4 $S_{4}$
S5 $S_{5}$

Solution

S1=−2 $S_{1} = - 2$
S4=−2+4+(−6)+8=4 $S_{4} = - 2 + 4 + (- 6) + 8 = 4$
S5=−2+4+(−6)+8+(−10)=−6 $S_{5} = - 2 + 4 + (- 6) + 8 + (- 10) = - 6$

Now Try Exercise 29.

Sigma Notation

The Greek letter Σ $Σ$ (sigma) can be used to denote a sum when the general term of a sequence is a formula. For example, the sum of the first four terms of the sequence 3, 5, 7, 9, …, 2k+1,… $2 k + 1, \dots$ can be named as follows, using what is called sigma notation, or summation notation:

\sum k = 1 4 (2 k + 1) .

$\sum_{k = 1}^{4} (2 k + 1) .$

This is read “the sum as k goes from 1 to 4 of 2k+1 $2 k + 1$ .” The letter k is called the index of summation. The index of summation might start at a number other than 1, and letters other than k can be used.

Example 4

Find and evaluate each of the following sums.

∑k=15k3 $\sum_{k = 1}^{5} k^{3}$
∑k=04(−1)k5k $\sum_{k = 0}^{4} {(- 1)}^{k} 5^{k}$
∑i=811(2+1i) $\sum_{i = 8}^{11} (2 + \frac{1}{i})$

Solution

We replace k with 1, 2, 3, 4, and 5. Then we add the results.

$\sum k = 1 5 k 3 = = = 13 + 23 + 33 + 43 + 53 1 + 8 + 27 + 64 + 125 225$ $\begin{array}{rcl} \sum_{k = 1}^{5} k^{3} & = & 1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} \\ = & 1 + 8 + 27 + 64 + 125 \\ = & 225 \end{array}$
$\sum k = 0 4 (- 1) k 5 k = = (- 1) 0 50 + (- 1) 1 51 + (- 1) 2 52 + (- 1) 3 53 + (- 1) 4 54 1 - 5 + 25 - 125 + 625 = 521$ $\begin{array}{rcl} \sum_{k = 0}^{4} {(- 1)}^{k} 5^{k} & = & {(- 1)}^{0} 5^{0} + {(- 1)}^{1} 5^{1} + {(- 1)}^{2} 5^{2} + {(- 1)}^{3} 5^{3} + {(- 1)}^{4} 5^{4} \\ = & 1 - 5 + 25 - 125 + 625 = 521 \end{array}$
$\sum i = 8 11 (2 + 1 i) = = (2 + 1 8) + (2 + 1 9) + (2 + 1 10) + (2 + 1 11) 8 1691 3960$ $\begin{array}{rcl} \sum_{i = 8}^{11} (2 + \frac{1}{i}) & = & (2 + \frac{1}{8}) + (2 + \frac{1}{9}) + (2 + \frac{1}{10}) + (2 + \frac{1}{11}) \\ = & 8 \frac{1691}{3960} \end{array}$

Now Try Exercise 33.

Example 5

Write sigma notation for each sum.

1+2+4+8+16+32+64 $1 + 2 + 4 + 8 + 16 + 32 + 64$
−2+4−6+8−10 $- 2 + 4 - 6 + 8 - 10$
x+x22+x33+x44+⋯ $x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \frac{x^{4}}{4} + \dots$

Solution

1+2+4+8+16+32+64 $1 + 2 + 4 + 8 + 16 + 32 + 64$

This is the sum of powers of 2, beginning with 20 $2^{0}$ , or 1, and ending with 26 $2^{6}$ , or 64. Sigma notation is ∑6k=02k $\sum_{k = 0}^{6} 2^{k}$ .
−2+4−6+8−10 $- 2 + 4 - 6 + 8 - 10$

Disregarding the alternating signs, we see that this is the sum of the first 5 even integers. Note that 2k is a formula for the kth positive even integer, and (−1)k=−1 ${(- 1)}^{k} = - 1$ when k is odd and (−1)k=1 ${(- 1)}^{k} = 1$ when k is even. Thus the general term is (−1)k(2k) ${(- 1)}^{k} (2 k)$ . The sum begins with k=1 $k = 1$ and ends with k=5 $k = 5$ , so sigma notation is ∑5k=1(−1)k(2k) $\sum_{k = 1}^{5} {(- 1)}^{k} (2 k)$ .
x+x22+x33+ x44+⋯ $x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \frac{x^{4}}{4} + \dots$

The general term is xk/k $x^{k} / k$ , beginning with k=1 $k = 1$ . This is also an infinite series. We use the symbol ∞ $\infty$ for infinity and write the series using sigma notation: ∑∞k=1(xk/k) $\sum_{k = 1}^{\infty} (x^{k} / k)$ .

Now Try Exercise 51.

Recursive Definitions

A sequence can be defined recursively or by using a recursion formula. Such a definition lists the first term, or the first few terms, and then describes how to determine the remaining terms from the given terms.

Example 6

Find the first 5 terms of the sequence defined by

a 1 = 5, a n + 1 = 2 a n - 3, for n \geq 1.

$a_{1} = 5, a_{n + 1} = 2 a_{n} - 3, for n \geq 1.$

Solution

We have

Now Try Exercise 61.

11.1 Exercise Set

In each of the following, the nth term of a sequence is given. Find the first 4 terms, a10 $a_{10}$ , and a15 $a_{15}$ .

1. an=4n−1 $a_{n} = 4 n - 1$
2. an=(n−1)(n−2)(n−3) $a_{n} = (n - 1) (n - 2) (n - 3)$
3. an=nn−1, n≥2 $a_{n} = \frac{n}{n - 1}, n \geq 2$
4. an=n2−1, n≥3 $a_{n} = n^{2} - 1, n \geq 3$
5. an=n2−1n2+1 $a_{n} = \frac{n^{2} - 1}{n^{2} + 1}$
6. an=(−12)n−1 $a_{n} = {(- \frac{1}{2})}^{n - 1}$
7. an=(−1)nn2 $a_{n} = {(- 1)}^{n} n^{2}$
11. an=(−1)n−1(3n−5) $a_{n} = {(- 1)}^{n - 1} (3 n - 5)$
9. an=5+(−2)n+12n $a_{n} = 5 + \frac{{(- 2)}^{n + 1}}{2^{n}}$
10. an=2n−1n2+2n $a_{n} = \frac{2 n - 1}{n^{2} + 2 n}$

Find the indicated term of the given sequence.

11. an=5n−6; a8 $a_{n} = 5 n - 6; a_{8}$
12. an=(3n−4)(2n+5); a7 $a_{n} = (3 n - 4) (2 n + 5); a_{7}$
13. an=(2n+3)2; a6 $a_{n} = {(2 n + 3)}^{2}; a_{6}$
14. an=(−1)n−1(4.6n−18.3); a12 $a_{n} = {(- 1)}^{n - 1} (4.6 n - 18.3); a_{12}$
15. an=5n2(4n−100); a11 $a_{n} = 5 n^{2} (4 n - 100); a_{11}$
16. an=(1+1n)2; a80 $a_{n} = {(1 + \frac{1}{n})}^{2}; a_{80}$
17. an=lnen; a67 $a_{n} = \ln e^{n}; a_{67}$
18. an=2−1000n; a100 $a_{n} = 2 - \frac{1000}{n}; a_{100}$

Predict the general term, or nth term, an $a_{n}$ , of the sequence. Answers may vary.

19. 2, 4, 6, 8, 10, …
20. 3, 9, 27, 81, 243, …
21. −2, 6, −18, 54,…
22. −2, 3, 8, 13, 18,…
23. 23, 34, 45, 56, 67,… $\frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \dots$
24. 2–√, 2,6–√, 22–√,10−−√,… $\sqrt{2}, 2, \sqrt{6}, 2 \sqrt{2}, \sqrt{10}, \dots$
25. 1⋅2, 2⋅3, 3⋅4, 4⋅5,… $1 \cdot 2, 2 \cdot 3, 3 \cdot 4, 4 \cdot 5, \dots$
26. −1, −4, −7, −10, −13,…
27. 0, log 10, log 100, log 1000,…
28. lne2, lne3, lne4, lne5,… $\ln e^{2}, \ln e^{3}, \ln e^{4}, \ln e^{5}, \dots$

Find the indicated partial sums for the sequence.

29. 1, 2, 3, 4, 5, 6, 7, …; S3 $S_{3}$ and S7 $S_{7}$
30. 1, −3, 5, −7, 9, −11,…; S2 $1, - 3, 5, - 7, 9, - 11, \dots; S_{2}$ and S5 $S_{5}$
31. 2, 4, 6, 8, …; S4 $S_{4}$ and S5 $S_{5}$
32. 1, 14, 19, 116, 125,…; S1 $1, \frac{1}{4}, \frac{1}{9}, \frac{1}{16}, \frac{1}{25}, \dots; S_{1}$ and S5 $S_{5}$

Find and evaluate the sum.

33. ∑k=1512k $\sum_{k = 1}^{5} \frac{1}{2 k}$
34. ∑i=1612i+1 $\sum_{i = 1}^{6} \frac{1}{2 i + 1}$
35. ∑i=062i $\sum_{i = 0}^{6} 2^{i}$
36. ∑k=472k−1−−−−−√ $\sum_{k = 4}^{7} \sqrt{2 k - 1}$
37. ∑k=710ln k $\sum_{k = 7}^{10} \ln k$
38. ∑k=14πk $\sum_{k = 1}^{4} π k$
39. ∑k=18kk+1 $\sum_{k = 1}^{8} \frac{k}{k + 1}$
40. ∑i=15i−1i+3 $\sum_{i = 1}^{5} \frac{i - 1}{i + 3}$
41. ∑i=15(−1)i $\sum_{i = 1}^{5} {(- 1)}^{i}$
42. ∑k=05(−1)k+1 $\sum_{k = 0}^{5} {(- 1)}^{k + 1}$
43. ∑k=18(−1)k+13k $\sum_{k = 1}^{8} {(- 1)}^{k + 1} 3 k$
44. ∑k=07(−1)k4k+1 $\sum_{k = 0}^{7} {(- 1)}^{k} 4^{k + 1}$
45. ∑k=062k2+1 $\sum_{k = 0}^{6} \frac{2}{k^{2} + 1}$
46. ∑i=110i(i+1) $\sum_{i = 1}^{10} i (i + 1)$
47. ∑k=05(k2−2k+3) $\sum_{k = 0}^{5} (k^{2} - 2 k + 3)$
48. ∑k=1101k(k+1) $\sum_{k = 1}^{10} \frac{1}{k (k + 1)}$
49. ∑i=0102i2i+1 $\sum_{i = 0}^{10} \frac{2^{i}}{2^{i} + 1}$
50. ∑k=03(−2)2k $\sum_{k = 0}^{3} {(- 2)}^{2 k}$

Write sigma notation. Answers may vary.

51. 5+10+15+20+25+⋯ $5 + 10 + 15 + 20 + 25 + \dots$
52. 7+14+21+28+35+⋯ $7 + 14 + 21 + 28 + 35 + \dots$
53. 2−4+8−16+32−64 $2 - 4 + 8 - 16 + 32 - 64$
54. 3+6+9+12+15 $3 + 6 + 9 + 12 + 15$
55. −12+23−34+45−56+67 $- \frac{1}{2} + \frac{2}{3} - \frac{3}{4} + \frac{4}{5} - \frac{5}{6} + \frac{6}{7}$
56. 112+122+132+142+152 $\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \frac{1}{5^{2}}$
57. 4−9+16−25+⋯+(−1)nn2 $4 - 9 + 16 - 25 + \dots + {(- 1)}^{n} n^{2}$
58. 9−16+25+⋯+(−1)n+1n2 $9 - 16 + 25 + \dots + {(- 1)}^{n + 1} n^{2}$
59. 11⋅2+12⋅3+13⋅4+14⋅5+⋯ $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \dots$
60. 11⋅22+12⋅32+13⋅42+14⋅52+⋯ $\frac{1}{1 \cdot 2^{2}} + \frac{1}{2 \cdot 3^{2}} + \frac{1}{3 \cdot 4^{2}} + \frac{1}{4 \cdot 5^{2}} + \dots$

Find the first 4 terms of the recursively defined sequence.

61. a1=4, an+1=1+1an $a_{1} = 4, a_{n + 1} = 1 + \frac{1}{a_{n}}$
62. a1=256, an+1=an−−√ $a_{1} = 256, a_{n + 1} = \sqrt{a_{n}}$
63. a1=6561, an+1=(−1)nan−−√ $a_{1} = 6561, a_{n + 1} = {(- 1)}^{n} \sqrt{a_{n}}$
64. a1=eQ, an+1=lnan $a_{1} = e^{Q}, a_{n + 1} = \ln a_{n}$
65. a1=2, a2=3, an+1=an+an−1 $a_{1} = 2, a_{2} = 3, a_{n + 1} = a_{n} + a_{n - 1}$
66. a1=−10, a2=8, an+1=an−an−1 $a_{1} = - 10, a_{2} = 8, a_{n + 1} = a_{n} - a_{n - 1}$
67. Compound Interest. Suppose that $1000 is invested at 6.2%, compounded annually. The value of the investment after n years is given by the sequence model

$a n = $1000 (1.062) n, n = 1, 2, 3, \dots .$ $a_{n} = $1000 {(1.062)}^{n}, n = 1, 2, 3, \dots .$
1. Find the first 10 terms of the sequence.
2. Find the value of the investment after 20 years.
68. Salvage Value. The value of a post-hole digger is $5200. Its salvage value each year is 75% of its value the year before. Give a sequence that lists the salvage value of the machine for each year of a 10-year period.
69. Wage Sequence. Adahy is paid $9.80 per hour for working at Red Freight Limited. Each year he receives a $1.10 hourly raise. Give a sequence that lists Adahy’s hourly wage over a 10-year period.
70. Bacteria Growth. Suppose that a single cell of bacteria divides into two every 15 min. Suppose that the same rate of division is maintained for 4 hr. Give a sequence that lists the number of cells after successive 15-min periods.
71. Fibonacci Sequence: Rabbit Population Growth. One of the most famous recursively defined sequences is the Fibonacci sequence. In 1202, the Italian mathematician Leonardo da Pisa, also called Fibonacci, proposed the following model for rabbit population growth. Suppose that every month each mature pair of rabbits in the population produces a new pair that begins reproducing after two months, and also suppose that no rabbits die. Beginning with one pair of newborn rabbits, the population can be modeled by the following recursively defined sequence:

$a 1 = 1, a 2 = 1, a n = a n - 1 + a n - 2, for n \geq 3,$ $a_{1} = 1, a_{2} = 1, a_{n} = a_{n - 1} + a_{n - 2}, for n \geq 3,$

where an $a_{n}$ is the total number of pairs of rabbits in month n. Find the first 7 terms of the Fibonacci sequence.

Skill Maintenance

Solve. [6.1], [6.3], [6.5], [6.6]

72. 3x−2y=3,2x+3y=−11 $\begin{matrix} 3 x - 2 y = 3, \\ 2 x + 3 y = - 11 \end{matrix}$
73. Harvesting Pumpkins. A total of 23,400 acres of pumpkins were harvested in Illinois and Ohio in 2012. The number of acres of pumpkins harvested in Ohio was 9000 fewer than the number of acres of pumpkins harvested in Illinois. (Source: U. S. Department of Agriculture) Find the number of acres of pumpkins harvested in Illinois and in Ohio in 2012.

Find the center and the radius of the circle with the given equation. [7.2]

74. x2+y2−6x+4y=3 $x^{2} + y^{2} - 6 x + 4 y = 3$
75. x2+y2+5x−8y=2 $x^{2} + y^{2} + 5 x - 8 y = 2$

Synthesis

Find the first 5 terms of the sequence, and then find S5 $S_{5}$ .

76. an=12nlog 1000n $a_{n} = \frac{1}{2^{n}} \log 1000^{n}$
77. an=in, i=−1−−−√ $a_{n} = i^{n}, i = \sqrt{- 1}$
78. an=ln (1⋅2⋅3⋯n) $a_{n} = \ln (1 \cdot 2 \cdot 3 \dots n)$

For each sequence, find a formula for Sn $S_{n}$ .

79. an=ln n $a_{n} = \ln n$
80. an=1n−1n+1 $a_{n} = \frac{1}{n} - \frac{1}{n + 1}$

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Table of Contents for 11.1 Sequences and Series

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Table of Contents for
11.1 Sequences and Series