5144 Demana Ch09pp699-790

9.4Sequences

What you’ll learn about

n Infinite Sequences

n Limits of Infinite Sequences

n Arithmetic and Geometric

Sequences

n Sequences and Graphing

Calculators

. . . and why

Infinite sequences, especially

those with finite limits, are

involved in some key concepts

of calculus.

Infinite Sequences

One of the most natural ways to study patterns in mathematics is to look at an ordered

progression of numbers, called a . Here are some examples of sequences:

1. 5, 10, 15, 20, 25

2. 2, 4, 8, 16, 32, . . . , 2k, . . .

3. 5}1

k}: k 5 1, 2, 3, . . .6

4. ha1, a2, a3, . . . , ak, . . .j, which is sometimes abbreviated hakj

The first of these is a , while the other three are

Notice that in (2) and (3) we were able to define a rule that gives the kth number in the

sequence (called the ) as a function of k. In (4) we do not have a rule, but

notice how we can use subscript notation sakd to identify the kth term of a “general”

infinite sequence. In this sense, an infinite sequence can be thought of as a function that

assigns a unique number sakd to each natural number k.

kth term

infinite sequences. finite sequence

sequence

732 CHAPTER 9 Discrete Mathematics

OBJECTIVE

Students will be able to express arith-

metic and geometric sequences explicitly

and recursively; they will also be able to

find limits of convergent sequences.

MOTIVATE

Ask students to find the 100th term of the

sequence 7, 10, 13, 16, … (304)

LESSON GUIDE

Day 1: Infinite Sequences; Limits of Infinite

Sequences

Day 2: Arithmetic and Geometric

Sequences; Sequences and Graphing

Calculators

AGREEMENT ON SEQUENCES

Since we will be dealing primarily with

infinite sequences in this book, the

word “sequence” will mean an infinite

sequence unless otherwise specified.

EXAMPLE 1 Defining a Sequence Explicitly

Find the first 6 terms and the 100th term of the sequence hakj in which ak 5 k2 2 1.

SOLUTION Since we know the kth term explicitly as a function of k, we need only

to evaluate the function to find the required terms:

a1 5 12 2 1 5 0, a2 5 3, a3 5 8, a4 5 15, a5 5 24, a6 5 35, and

a100 5 1002 2 1 5 9999.

Now try Exercise 1.

EXAMPLE 2 Defining a Sequence Recursively

Find the first 6 terms and the 100th term for the sequence defined recursively by the

conditions:

b1 5 3

bn 5 bn21 1 2 for all n . 1.

Explicit formulas are the easiest to work with, but there are other ways to define

sequences. For example, we can specify values for the first term (or terms) of a

sequence, then define each of the following terms by a formula relating it

to previous terms. Example 2 shows how this is done.

recursively

continued

5144_Demana_Ch09pp699-790 1/13/06 3:06 PM Page 732

SECTION 9.4 Sequences 733

ALERT

In giving a recursive definition for a

sequence, some students will give only the

recursive formula. Point out in Example 2

that the formula bn 5 bn21 1 2 could

define many different sequences—but only

one sequence that begins with b1 5 3.

TEACHING NOTE

Consistent with the philosophy of this

book, we are using limit notation here

because it is intuitively clear to students.

It is easy to push any discussion of limits

beyond the realm of intuitive clarity, but

we do not recommend it in this course.

For the same reason, we are using certain

properties of limits here without proving

or stating them as theorems.

SOLUTION We proceed one term at a time, starting with b1 5 3 and obtaining each

succeeding term by adding 2 to the term just before it:

b1 5 3

b2 5 b1 1 2 5 5

b3 5 b2 1 2 5 7

etc.

Eventually it becomes apparent that we are building the sequence of odd natural num-

bers beginning with 3:

h3, 5, 7, 9, . . .j.

The 100th term is 99 terms beyond the first, which means that we can get there quickly

by adding 99 2’s to the number 3:

b100 5 3 1 99 3 2 5 201.

Now try Exercise 5.

Limits of Infinite Sequences

Just as we were concerned with the end behavior of functions, we will also be con-

cerned with the end behavior of sequences.

EXAMPLE 3 Finding Limits of Sequences

Determine whether the sequence converges or diverges. If it converges, give the limit.

(a) }1

1}, }

1

2}, }

1

3}, }

1

4}, . . . , }

1

n}, . . .

(b) }2

1}, }

3

2}, }

4

3}, }

5

4}, . . .

(c) 2, 4, 6, 8, 10, . . .

(d) 21, 1, 21, 1, . . . , (21)n, . . .

SOLUTION

(a) limny`

}1

n} 5 0, so the sequence converges to a limit of 0.

(b) Although the nth term is not explicitly given, we can see that an 5 }n 1

n

1}.

limny`

}n 1

n

1} 5 lim

ny` (1 1 }1

n} ) 5 1 1 0 5 1. The sequence converges to a limit of 1.

(c) This time we see that an 5 2n. Since limny`

2n 5 `, the sequence diverges.

(d) This sequence oscillates forever between two values and hence has no limit. The

sequence diverges.

Now try Exercise 13.

DEFINITION Limit of a Sequence

Let hanj be a sequence of real numbers, and consider limny`

an. If the limit is

a finite number L, the sequence converges and L is the limit of the sequence. If the

limit is infinite or nonexistent, the sequence diverges.



DEFINITION Arithmetic Sequence

A sequence hanj is an if it can be written in the form

ha, a 1 d, a 1 2d, . . . , a 1 sn 2 1dd, . . .j for some constant d.

The number d is called the .

Each term in an arithmetic sequence can be obtained recursively from its preceding

term by adding d:

an 5 an21 1 d (for all n $ 2).

common difference

arithmetic sequence

TEACHING NOTE

Emphasize that an arithmetic or geometric

sequence {an} can be defined in either of

two ways: (1) explicitly, in terms of the

variable n, or (2) recursively, in terms of the

previous term an21.

It might help to review the rules for finding the end behavior asymptotes of rational

functions (page 240 in Section 2.6) because those same rules apply to sequences that

are rational functions of n, as in Example 4.

Arithmetic and Geometric Sequences

There are all kinds of rules by which we can construct sequences, but two particular types

of sequences dominate in mathematical applications: those in which pairs of successive

terms all have a common difference ( sequences), and those in which pairs of

successive terms all have a common quotient, or ratio ( sequences). We will

take a closer look at these in this section.

geometric

arithmetic

PRONUNCIATION TIP

The word “arithmetic” is probably

more familiar to you as a noun, refer-

ring to the mathematics you studied

in elementary school. In this word,

the second syllable (“rith”) is accent-

ed. When used as an adjective, the

third syllable (“met”) gets the accent.

(For the sake of comparison, a similar

shift of accent occurs when going

from the noun “analysis” to the

adjective “analytic”)

EXAMPLE 5 Defining Arithmetic Sequences

For each of the following arithmetic sequences, find (a) the common difference,

(b) the tenth term, (c) a recursive rule for the nth term, and (d) an explicit rule for the

nth term.

(1) 26, 22, 2, 6, 10, . . .

(2) ln 3, ln 6, ln 12, ln 24, . . .

EXAMPLE 4 Finding Limits of Sequences

Determine whether the sequence converges or diverges. If it converges, give the limit.

(a) 5}n3

1

n

1}6 (b) 5}n3

5

1

n2

1}6 (c) 5}nn

3

2

1

1

2

n}6

SOLUTION

(a) Since the degree of the numerator is the same as the degree of the denominator,

the limit is the ratio of the leading coefficients.

Thus limny`

}n

3

1

n

1} 5 }

3

1} 5 3. The sequence converges to a limit of 3.

(b) Since the degree of the numerator is less than the degree of the denominator, the

limit is zero. Thus limny`

}n3

5

1

n2

1} 5 0. The sequence converges to 0.

(c) Since the degree of the numerator is greater than the degree of the denominator,

the limit is infinite. Thus limny`

}n

n

3

2

1

1

2

n} is infinite. The sequence diverges.


continued



SOLUTION

(1) (a) The difference between successive terms is 4.

(b) a10 5 26 1 (10 2 1)(4) 5 30

(c) The sequence is defined recursively by a1 5 26 and an 5 an21 1 4 for all

n $ 2.

(d) The sequence is defined explicitly by an 5 26 1 (n 2 1)(4) 5 4n 2 10.

(2) (a) This sequence might not look arithmetic at first, but

ln 6 2 ln 3 5 ln }6

3} 5 ln 2 (by a law of logarithms) and the difference

between successive terms continues to be ln 2.

(b) a10 5 ln 3 1 (10 2 1)ln 2 5 ln 3 19 ln 2 5 ln(3 • 29) 5 ln 1536

(c) The sequence is defined recursively by a1 5 ln 3 and an 5 an21 1 ln 2

for all n $ 2.

(d) The sequence is defined explicitly by an 5 ln 3 1 (n 2 1)ln 2

5 ln (3 • 2n21). Now try Exercise 21.

DEFINITION Geometric Sequence

A sequence hanj is a if it can be written in the form

ha, a • r, a • r2, . . . , a • rn21, . . .j for some nonzero constant r.

The number r is called the .

Each term in a geometric sequence can be obtained recursively from its preceding

term by multiplying by r:

an 5 an21 • r (for all n $ 2).

common ratio

geometric sequence

EXAMPLE 6 Defining Geometric Sequences

For each of the following geometric sequences, find (a) the common ratio, (b) the tenth

term, (c) a recursive rule for the nth term, and (d) an explicit rule for the nth term.

(1) 3, 6, 12, 24, 48, . . .

(2) 1023, 1021, 101, 103, 105, . . .

SOLUTION

(1) (a) The ratio between successive terms is 2.

(b) a10 5 3 • 21021 5 3 • 29 5 1536

(c) The sequence is defined recursively by a1 5 3 and an 5 2an21

for n $ 2.

(d) The sequence is defined explicitly by an 5 3 • 2n21.

continued



EXAMPLE 7 Constructing Sequences

The second and fifth terms of a sequence are 3 and 24, respectively. Find explicit and

recursive formulas for the sequence if it is (a) arithmetic and (b) geometric.

SOLUTION

(a) If the sequence is arithmetic, then a2 5 a1 1 d 5 3 and a5 5 a1 1 4d 5 24.

Subtracting, we have

(a1 1 4d) 2 (a1 1 d) 5 24 2 3

3d 5 21

d 5 7

Then a1 1 d 5 3 implies a1 5 24.

The sequence is defined explicitly by an 5 24 1 (n 2 1) • 7, or an 5 7n 2 11.

The sequence is defined recursively by a1 5 24 and an 5 an21 1 7 for n $ 2.

(b) If the sequence is geometric, then a2 5 a • r1 5 3 and a5 5 a • r4 5 24. Dividing,

we have

}a

a

1

1

•

•

r

r

4

1} 5 }

2

3

4}

r3 5 8

r 5 2

Then a1 • r1 5 3 implies a1 5 1.5.

The sequence is defined explicitly by an 5 1.5(2)n21, or an 5 3(2)n22.

The sequence is defined recursively by a1 5 1.5 and an 5 2 • an21


(2) (a) Applying a law of exponents, 5 10212(23) 5 102, and the ratio between

(a) successive terms continues to be 102.

(b) a10 5 1023 • (102)1021 5 1023118 5 1015

(c) The sequence is defined recursively by a1 5 1023 and an 5 102an 2 1 for n $ 2.

(d) The sequence is defined explicitly by an 5 1023(102)n21 5 102312n22 5 102n25.


1021

}1023



Sequences and Graphing Calculators

As with other kinds of functions, it helps to be able to represent a sequence geometri-

cally with a graph. There are at least two ways to obtain a sequence graph on a graph-

ing calculator. One way to graph explicitly defined sequences is as scatter plots of

points of the form sk, akd. A second way is to use the sequence graphing mode on a

graphing calculator.

EXAMPLE 8 Graphing a Sequence Defined Explicitly

Produce on a graphing calculator a graph of the sequence hakj in which

ak 5 k2 2 1.

Method 1 (Scatter Plot)

The command seqsK, K, 1, 10dy L1 puts the first 10 natural numbers in list L1. (You

could change the 10 if you wanted to graph more or fewer points.)

The command L12

2 1y L2 puts the corresponding terms of the sequence in list L2.

A scatter plot of L1, L2 produces the graph in Figure 9.7a.

Method 2 (Sequence Mode)

With your calculator in Sequence mode, enter the sequence ak 5 k2 2 1 in the Y 5

list as u(n) 5 n2 2 1 with nMin 5 1, nMax 5 10, and u(nMin) 5 0. (You could

change the 10 if you wanted to graph more or fewer points.) Figure 9.7b shows the

graph in the same window as Figure 9.7a.

FIGURE 9.7 The sequence ak 5 k2 2 1 graphed (a) as a scatter plot and (b) using the

sequence graphing mode. Tracing along the points gives values of ak for k 5 1, 2, 3, . . .

(Example 8)


[–1, 15] by [–10, 100]

(b)

X=6 Y=35

n=6

1

X=6 Y=35

[–1, 15] by [–10, 100]

(a)

EXAMPLE 9 Generating a Sequence with a Calculator

Using a graphing calculator, generate the specific terms of the following sequences:

(a) (Explicit) ak 5 3k 2 5 for k 5 1, 2, 3, . . .

(b) (Recursive) a1 5 22 and an 5 an21 1 3 for n 5 2, 3, 4, . . .

SEQUENCE GRAPHING

Most graphers enable you to graph in

“sequence mode.” Check your owner’s

manual to see how to use this mode.

FOLLOW-UP

Ask students to discuss the relative advan-

tages and disadvantages of graphing

sequences in scatter plots or in sequence

mode.

ASSIGNMENT GUIDE

Day 1: Ex. 1–20

Day 2: Ex. 21–36, multiples of 3;

Ex. 43–48

COOPERATIVE LEARNING

Group Activity: Ex. 41, 42

NOTES ON EXERCISES

Ex. 41, 42 are examples of "mathemagic."

Ex. 43–48 provide practice for standardized

tests.

ONGOING ASSESSMENT

Self-Assessment: Ex. 1, 5, 13, 15, 21, 25,

29, 33

Embedded Assessment: Ex. 49, 53, 54

continued



FIGURE 9.9 Typing these two commands

(on the left of the viewing screen) will

generate the terms of the recursively defined

sequence with a1 5 22 and an 5 an21 1 3

(Example 9b).

–2

Ans+3

4

1

–2

10

7

FIGURE 9.10 The two commands on the left will generate the Fibonacci sequence as the

ENTER key is pressed repeatedly.

The Fibonacci sequence can be defined recursively using three statements.

0 A:1 B

A+B C:A B:C A

1

2

3

1

1

The Fibonacci Sequence

The Fibonacci sequence can be defined recursively by

a1 5 1

a2 5 1

an 5 an22 1 an21

for all positive integers n $ 3.

FIBONACCI NUMBERS

The numbers in the Fibonacci

sequence have fascinated profession-

al and amateur mathematicians alike

since the thirteenth century. Not only

is the sequence, like Pascal’s triangle,

a rich source of curious internal pat-

terns, but the Fibonacci numbers

seem to appear everywhere in

nature. If you count the leaflets on a

leaf, the leaves on a stem, the whorls

on a pine cone, the rows on an ear of

corn, the spirals in a sunflower, or the

branches from a trunk of a tree, they

tend to be Fibonacci numbers. (Check

phyllotaxy in a biology book.)

SOLUTION

(a) On the home screen, type the two commands shown in Figure 9.8. The calculator will

then generate the terms of the sequence as you push the ENTER key repeatedly.

(b) On the home screen, type the two commands shown in Figure 9.9. The first com-

mand gives the value of a1. The calculator will generate the remaining terms of

the sequence as you push the ENTER key repeatedly.

Notice that these two definitions generate the very same sequence!

Now try Exercises 1 and 5 on your calculator.

A recursive definition of an can be made in terms of any combination of preceding

terms, just as long as those preceding terms have already been determined. A famous

example is the , named for Leonardo of Pisa (ca. 1170–1250),

who wrote under the name Fibonacci. You can generate it with the two commands

shown in Figure 9.10.

Fibonacci sequence

FIGURE 9.8 Typing these two commands

(on the left of the viewing screen) will

generate the terms of the explicitly defined

sequence ak 5 3k 2 5. (Example 9a)

0 K

K+1 K:3K–5

1

–2

0

7

4



QUICK REVIEW 9.4 (For help, see Section P.1.)

In Exercises 1 and 2, evaluate each expression when a 5 3, d 5 4,

and n 5 5.

1. a 1 sn 2 1dd 19 2. }n

2} s2a 1 sn 2 1ddd 55

In Exercises 3 and 4, evaluate each expression when a 5 5, r 5 4,

and n 5 3.

3. a • rn21 80 4. 105as1 2 rnd}}

1 2 r

In Exercises 5–10, find a10.

5. ak 5 }k 1

k

1} 10/11 6. ak 5 5 1 sk 2 1d3 32

7. ak 5 5 • 2k21 2560 8. ak 5 }4

3} (}

1

2})

k21

1/384

9. ak 5 32 2 ak21 and a9 5 17 15

10. ak 5 25/256k2

}2k

SECTION 9.4 EXERCISES

In Exercises 1–4, find the first 6 terms and the 100th term of the

explicitly-defined sequence.

1. un 5 }n 1

n

1} 2. vn 5 }

n 1

4

2}

3. cn 5 n3 2 n 4. dn 5 n2 2 5n

In Exercises 5–10, find the first 4 terms and the eighth term of the

recursively-defined sequence.

5. a1 5 8 and an 5 an21 2 4, for n $ 2 8, 4, 0, 24; 220

6. u1 5 23 and uk11 5 uk 1 10, for k $ 1 23, 7, 17, 27; 67

7. b1 5 2 and bk11 5 3bk, for k $ 1 2, 6, 18, 54; 4374

8. v1 5 0.75 and vn 5 s22dvn21, for n $ 2

9. c1 5 2, c2 5 21, and ck12 5 ck 1 ck11, for k $ 1

10. c1 5 22, c2 5 3, and ck 5 ck22 1 ck21, for k $ 3

In Exercises 11–20, determine whether the sequence converges or

diverges. If it converges, give the limit.

11. 1, 4, 9, 16, . . . , n2, . . . Diverges

12. }1

2}, }

1

4}, }

1

8}, }

1

1

6}, . . . , }

2

1n}, . . . Converges to 0

13. }1

1}, }

1

4}, }

1

9}, }

1

1

6}, . . . Converges to 0

14. h3n 2 1j Diverges

15. 5}32n

2

2

3

1

n}6 Converges to 21

16. 5}2nn

1

2

1

1}6 Converges to 2

17. h(0.5)nj Converges to 0

18. h(1.5)nj Diverges

19. a1 5 1 and an11 5 an 1 3 for n $ 1 Diverges

20. u1 5 1 and un11 5 }u

3n} for n $ 1 Converges to 0

In Exercises 21–24, the sequences are arithmetic. Find

(a) the common difference,

(b) the tenth term,

(c) a recursive rule for the nth term, and

(d) an explicit rule for the nth term.

21. 6, 10, 14, 18, . . . 22. 24, 1, 6, 11, . . .

23. 25, 22, 1, 4, . . . 24. 27, 4, 15, 26, . . .

In Exercises 25–28, the sequences are geometric. Find

(a) the common ratio,

(b) the eighth term,

(c) a recursive rule for the nth term, and

(d) an explicit rule for the nth term.

25. 2, 6, 18, 54, . . . 26. 3, 6, 12, 24, . . .

27. 1, 22, 4, 28, 16, . . . 28. 22, 2, 22, 2, . . .

29. The fourth and seventh terms of an arithmetic sequence are 28

and 4, respectively. Find the first term and a recursive rule for the

nth term. a1 5 220; an 5 an21 1 4 for n $ 2

30. The fifth and ninth terms of an arithmetic sequence are 25 and

217, respectively. Find the first term and a recursive rule for the

nth term. a1 5 7; an 5 an21 2 3 for n $ 2

31. The second and eighth terms of a geometric sequence are 3 and

192, respectively. Find the first term, common ratio, and an

explicit rule for the nth term.

32. The third and sixth terms of a geometric sequence are 275 and

29375, respectively. Find the first term, common ratio, and an

explicit rule for the nth term.

In Exercises 33–36, graph the sequence.

33. an 5 2 2 }1

n} 34. bn 5 Ïnw 2 3

35. cn 5 n2 2 5 36. dn 5 3 1 2n


37. Rain Forest Growth The bungy-

bungy tree in the Amazon rain forest

grows an average 2.3 cm per week. Write

a sequence that represents the weekly

height of a bungy-bungy over the course of

1 year if it is 7 meters tall today. Display

the first four terms and the last two terms.

38. Half-Life (See Section 3.2) Thorium-232 has a half-life of 14 bil-

lion years. Make a table showing the half-life decay of a sample of

Thorium-232 from 16 grams to 1 gram; list the time (in years, start-

ing with t 5 0) in the first column and the mass (in grams) in the sec-

ond column. Which type of sequence is each column of the table?

39. Arena Seating The first row of seating in section J of the

Athena Arena has 7 seats. In all, there are 25 rows of seats in sec-

tion J, each row containing two more seats than the row preceding

it. How many seats are in section J? 775

40. Patio Construction Pat designs a patio with a trapezoid-

shaped deck consisting of 16 rows of congruent slate tiles. The

numbers of tiles in the rows form an arithmetic sequence. The

first row contains 15 tiles and the last row contains 30 tiles. How

many tiles are used in the deck? 360

41. Group Activity Pair up with a partner to create a sequence

recursively together. Each of you picks five random digits from 1

to 9 (with repetitions, if you wish). Merge your digits to make a

list of ten. Now each of you constructs a ten-digit number using

exactly the numbers in your list.

Let a1 5 the (positive) difference between your two numbers.

Let an11 5 the sum of the digits of an for n $ 1.

This sequence converges, since it is eventually constant. What is

the limit? (Remember, you can check your answer in the back of

the book.)

42. Group Activity Here is an interesting recursively defined word

sequence. Join up with three or four classmates and, without

telling it to the others, pick a word from this sentence. Then, with

care, count the letters in your word. Move ahead that many words

in the text to come to a new word. Count the letters in the new

word. Move ahead again, and so on. When you come to a point

when your next move would take you out of this problem, stop.

Share your last word with your friends. Are they all the same?

Standardized Test Questions43. True or False If the first two terms of a geometric sequence are

negative, then so is the third. Justify your answer.

44. True or False If the first two terms of an arithmetic sequence

are positive, then so is the third. Justify your answer.

You may use a graphing calculator when solving Exercises 45–48.

45. Multiple Choice The first two terms of an arithmetic sequence

are 2 and 8. The fourth term is A

(A) 20. (B) 26. (C) 64. (D) 128. (E) 256.

46. Multiple Choice Which of the following sequences is

divergent? B

(A) 5}n 1

n

100}6 (B) hÏnwj (C) hp2nj (D) 5}2n

n

1

1

1

2}6 (E) hn22j

47. Multiple Choice A geometric sequence hanj begins 2, 6, . . . .

What is }a

a6

2

}? E

(A) 3 (B) 4 (C) 9 (D) 12 (E) 81

48. Multiple Choice Which of the following rules for n $ 1 will

define a geometric sequence if a1 Þ 0? C

(A) an11 5 an 1 3 (B) an11 5 an 2 3 (C) an11 5 an 4 3

(D) an11 5 a3n (E) an11 5 an • 3n21

Explorations49. Rabbit Populations Assume that 2 months after birth, each

male-female pair of rabbits begins producing one new male-

female pair of rabbits each month. Further assume that the rabbit

colony begins with one newborn male-female pair of rabbits and

no rabbits die for 12 months. Let an represent the number of pairs

of rabbits in the colony after n 2 1 months.

(a) Writing to Learn Explain why a1 5 1, a2 5 1, and a3 5 2.

(b) Find a4, a5, a6, . . . , a13.

(c) Writing to Learn Explain why the sequence hanj, 1 # n #

13, is a model for the size of the rabbit colony

for a 1-year period.

50. Fibonacci Sequence Compute the first seven terms of the

sequence whose nth term is

an 5 ( )n

2 ( )n

.

How do these seven terms compare with the first seven terms of

the Fibonacci sequence?

51. Connecting Geometry and Sequences In the following

sequence of diagrams, regular polygons are inscribed in unit cir-

cles with at least one side of each polygon perpendicular to the

positive x-axis.

1 2 Ï5w}

2

1}Ï5w

1 1 Ï5w}

2

1}Ï5w


(a) (b)

(c) (d)

y y

y y

x x

x x

1 1

1 1


(a) Prove that the perimeter of each polygon in the sequence is

given by an 5 2n sin sp/nd, where n is the number of sides in

the polygon.

(b) Investigate the value of an for n 5 10, 100, 1000, and

10,000. What conclusion can you draw? an → 2p as n → `

52. Recursive Sequence The population of Centerville was

525,000 in 1992 and is growing annually at the rate of 1.75%.

Write a recursive sequence hPnj for the population. State the first

term P1 for your sequence.

53. Writing to Learn If hanj is a geometric sequence with all posi-

tive terms, explain why hlog anj must be arithmetic.

54. Writing to Learn If hbnj is an arithmetic sequence, explain

why h10bnj must be geometric.

Extending the Ideas55. A Sequence of Matrices Write out the first seven terms of the

“geometric sequence” with the first term the matrix

f1 1g and the common ratio the matrix 3 4. How is this

sequence of matrices related to the Fibonacci sequence?

56. Another Sequence of Matrices Write out the first seven

terms of the “geometric sequence” which has for its first term the

matrix f1 ag and for its common ratio the matrix 3 4.How is this sequence of matrices related to the arithmetic

sequence?

1 d

0 1

0 1

1 1


52. P1 5 525,000 and Pn 5 1.0175Pn 2 1, n $ 2 55. a1 5 [1 1], a2 5 [1 2], a3 5 [2 3], a4 5 [3 5], a5 5 [5 8], a6 5 [8 13], a7 5 [13 21].

The entries in the terms of this sequence are successive pairs of terms from the Fibonacci sequence. 56. Each matrix has the form [1 a 1 (n 2 1)d ], so the

second entries of this geometric sequence of matrices form an arithmetic sequence with first term a and common difference d.



9.5Series

What you’ll learn about

n Summation Notation

n Sums of Arithmetic and

Geometric Sequences

n Infinite Series

n Convergence of Geometric

Series

. . . and why

Infinite series are at the heart

of integral calculus.

Summation Notation

We want to look at the formulas for summing the terms of arithmetic and geometric

sequences, but first we need a notation for writing the sum of an indefinite number

of terms. The capital Greek letter sigma sSd provides our shorthand notation for a

“summation.”

DEFINITION Summation Notation

In , the sum of the terms of the sequence ha1, a2, . . . , anj is

denoted

^n

k51

ak

which is read “the sum of ak from k 5 1 to n.”

The variable k is called the .index of summation

summation notation

Although you probably computed them correctly, there is more going on in number 4

and number 5 in the above exploration than first meets the eye. We will have more to

say about these “infinite” summations toward the end of this section.

Sums of Arithmetic and Geometric Sequences

One of the most famous legends in the lore of mathematics concerns the German

mathematician Karl Friedrich Gauss (1777–1855), whose mathematical talent was

apparent at a very early age. One version of the story has Gauss, at age ten, being

EXPLORATION EXTENSIONS

Ask the students to speculate about the

value of #3 if 12 is replaced by a larger even

integer or a larger odd integer.

EXPLORATION 1 Summing with Sigma

Sigma notation is actually even more versatile than the definition above sug-

gests. See if you can determine the number represented by each of the follow-

ing expressions.

1. ^5

k51

3k 45 2. ^8

k55

k2 174 3. ^12

n50

cos snpd 1 4. ^`

n51

sin snpd 0 5. ^`

k51

}1

3

0k} }

1

3}

(If you’re having trouble with number 5, here’s a hint: Write the sum as a

decimal!)

SUMMATIONS ON A CALCULATOR

If you think of summations as sum-

ming sequence values, it is not hard

to translate sigma notation into cal-

culator syntax. Here, in calculator

syntax, are the first three summa-

tions in Exploration 1. (Don’t try

these on your calculator until you

have first figured out the answers

with pencil and paper.)

1. sum (seq (3K, K, 1, 5))

2. sum (seq (K^2, K, 5, 8))

3. sum (seq (cos (Np), N, 0, 12))


SECTION 9.5 Series 743

OBJECTIVE

Students will be able to use sigma notation

and find finite sums of terms in arithmetic

and geometric sequences; they will also be

able to find sums of convergent geometric

series.

MOTIVATE

Have the students speculate what the sum of

1/2 1 1/4 1 1/8 1 · · · might be. (1)

LESSON GUIDE

Day 1: Summation Notation; Sums of

Arithmetic and Geometric Sequences

Day 2: Infinite Series; Convergence of

Geometric Series

in a class that was challenged by the teacher to add up all the numbers from 1 to

100. While his classmates were still writing down the problem, Gauss walked to the

front of the room to present his slate to the teacher. The teacher, certain that Gauss

could only be guessing, refused to look at his answer. Gauss simply placed it face

down on the teacher’s desk, declared “There it is,” and returned to his seat. Later,

after all the slates had been collected, the teacher looked at Gauss’s work, which

consisted of a single number: the correct answer. No other student (the legend goes)

got it right.

The important feature of this legend for mathematicians is how the young Gauss got the

answer so quickly. We’ll let you reproduce his technique in Exploration 2.

EXPLORATION 2 Gauss’s Insight

Your challenge is to find the sum of the natural numbers from 1 to 100 with-

out a calculator.

1. On a wide piece of paper, write the sum 1 1 2 1 3 1 . . . 1 99 1 100

“1 1 2 1 3 1 ? ? ? 1 98 1 99 1 100.”

2. Underneath this sum, write the sum 100 1 99 1 98 1 . . . 1 2 1 1

“100 1 99 1 98 1 ? ? ? 1 3 1 2 1 1.”

3. Add the numbers two-by-two in vertical columns and notice that you get

the same identical sum 100 times. What is it? 101

4. What is the sum of the 100 identical numbers referred to in part 3? 10, 100

5. Explain why half the answer in part 4 is the answer to the challenge. Can

you find it without a calculator? The sum in 4 involves two copies of the same pro-

gression, so it doubles the sum of the progression. The answer is 5050.

EXPLORATION EXTENSIONS

Have the students write the series

1 1 2 1 3 1 . . . 1 98 1 99 1 100 and

100 1 99 1 98 1 . . . 1 3 1 2 1 1 in

sigma notation. Then discuss the properties

of summation

(1) ^n

k51

ak 1 ^n

k51

bk 5 ^n

k51

(ak 1 bk) and

(2) ^n

k51

cak 5 c ^n

k51

ak

(where c is a constant) and how these prop-

erties relate to Gauss’s insight.If this story is true, then the youthful Gauss had discovered a fact that his elders knew

about arithmetic sequences. If you write a finite arithmetic sequence forward on one line

and backward on the line below it, then all the pairs stacked vertically sum to the same

number. Multiplying this number by the number of terms n and dividing by 2 gives us a

shortcut to the sum of the n terms. We state this result as a theorem.

THEOREM Sum of a Finite Arithmetic Sequence

Let ha1, a2, a3, . . . , anj be a finite arithmetic sequence with common difference d.

Then the sum of the terms of the sequence is

^n

k51

ak 5 a1 1 a2 1 ? ? ? 1 an

5 n (}a1 1

2

an} )

5 }n

2} s2a1 1 sn 2 1dd d



Proof

We can construct the sequence forward by starting with a1 and adding d each time, or we

can construct the sequence backward by starting at an and subtracting d each time. We

thus get two expressions for the sum we are looking for:

^n

k51

ak 5 a1 1 sa1 1 dd 1 sa1 1 2dd 1 ? ? ? 1 sa1 1 sn 2 1ddd

^n

k51

ak 5 an 1 san 2 dd 1 san 2 2dd 1 ? ? ? 1 san 2 sn 2 1ddd

Summing vertically, we get

2 ^n

k51

ak¬5 sa1 1 and 1 sa1 1 and 1 ? ? ? 1 sa1 1 and

2 ^n

k51

ak¬5 nsa1 1 and

^n

k51

ak¬5 n (}a1 1

2

an})

If we substitute a1 1 sn 2 1dd for an, we get the alternate formula

^n

k51

ak 5 }n

2} s2a1 1 sn 2 1ddd.

EXAMPLE 1 Summing the Terms of an Arithmetic

Sequence

A corner section of a stadium has 8 seats along the front row. Each successive row

has two more seats than the row preceding it. If the top row has 24 seats, how many

seats are in the entire section?

SOLUTION The numbers of seats in the rows form an arithmetic sequence with

a1 5 8, an 5 24, and d 5 2.

Solving an 5 a1 1 sn 2 1dd, we find that

24 5 8 1 sn 2 1ds2d

16 5 sn 2 1ds2d

8 5 n 2 1

n 5 9

Applying the Sum of a Finite Arithmetic Sequence Theorem, the total number of seats

in the section is 9s8 1 24d/2 5 144.

We can support this answer numerically by computing the sum on a calculator:

sum(seqs8 1 sN 2 1d2, N, 1, 9d 5 144.

Now try Exercise 7.



As you might expect, there is also a convenient formula for summing the terms of a

finite geometric sequence.

EXAMPLE 2 Summing the Terms of a Geometric

Sequence

Find the sum of the geometric sequence 4, 24/3, 4/9, 24/27, . . . , 4(21/3)10.

SOLUTION We can see that a1 5 4 and r 5 21/3. The nth term is 4(21/3d10,

which means that n 5 11. (Remember that the exponent on the nth term is n 2 1, not

n.) Applying the Sum of a Finite Geometric Sequence Theorem, we find that

^11

n51

4 (2}1

3} )

n21

5 < 3.000016935.

We can support this answer by having the calculator do the actual summing:

sumsseqs4s21/3d^sN 2 1d, N, 1, 11d 5 3.000016935. Now try Exercise 13.

4s1 2 s21/3d11d}}

1 2 s21/3d

THEOREM Sum of a Finite Geometric Sequence

Let ha1, a2, a3, . . . , anj be a finite geometric sequence with common ratio r Þ 1.

Then the sum of the terms of the sequence is

^n

k51

ak 5 a1 1 a2 1 ? ? ? 1 an

5 }a1(

1

1

2

2

r

rn)}

Proof

Because the sequence is geometric, we have

^n

k51

ak¬5 a1 1 a1 • r 1 a1 • r2 1 ? ? ? 1 a1 • rn21.

Therefore,

r • ^n

k51

ak¬5 a1 • r 1 a1 • r2 1 ? ? ? 1 a1 • rn21 1 a1 • rn.

If we now subtract the lower summation from the one above it, we have (after elimi-

nating a lot of zeros):

( ^n

k51

ak) 2 r • ( ^n

k51

ak)¬5 a1 2 a1 • rn

( ^n

k51

ak)s1 2 rd¬5 a1s1 2 rnd

^n

k51

ak¬5 }a1(

1

1

2

2

r

rn)}



FOLLOW-UP

Ask if an infinite arithmetic series can ever

converge. (Only if it is a constant series of

0’s, which is not arithmetic in the usual

sense since d 5 0.)

ASSIGNMENT GUIDE

Day 1: Ex. 3–24, multiples of 3; Ex. 35, 37

Day 2: Ex. 25–32; Ex. 41–46

COOPERATIVE LEARNING

Group Activity: Ex. 39, 40

NOTES ON EXERCISES

Ex. 1–34 are standard summation and series

problems, to provide practice with the for-

mulas.

Ex. 41–46 provide practice for standardized

tests.

ONGOING ASSESSMENT

Self-Assessment: Ex. 7, 13, 23, 25, 31

Embedded Assessment: Ex. 35, 47, 49

DEFINITION Infinite Series

An is an expression of the form

^`

n51

an 5 a1 1 a2 1 ? ? ? 1an 1 ? ? ?.

infinite series

As one practical application of the Sum of a Finite Geometric Sequence Theorem, we

will tie up a loose end from Section 3.6, wherein you learned that the future value FV

of an ordinary annuity consisting of n equal periodic payments of R dollars at an inter-

est rate i per compounding period (payment interval) is

FV 5 R}s1 1 i

i

dn 2 1}.

We can now consider the mathematics behind this formula. The n payments remain in

the account for different lengths of time and so earn different amounts of interest. The

total value of the annuity after n payment periods (see Example 8 in Section 3.6) is

FV 5 R 1 Rs1 1 id 1 Rs1 1 id2 1 ? ? ? 1 Rs1 1 idn21.

The terms of this sum form a geometric sequence with first term R and common ratio

s1 1 id. Applying the Sum of a Finite Geometric Sequence Theorem, the sum of the n

terms is

FV 5}R(

1

1

2

2

(

(

1

1

1

1

i

i

ddn)

}

5 R}1 2 s

2

1

i

1 idn

}

5 R}(1 1 i

i

)n 2 1}

Infinite Series

If you change the “11” in the calculator sum in Example 2 to higher and higher num-

bers, you will find that the sum approaches a value of 3. This is no coincidence. In the

language of limits,

limn→`

^n

k51

4 (2}1

3} )

k21

5 limn→`

5 }4s1

4/

2

3

0d} Since lim

n→∞

(21/3)n 5 0.

5 3

This gives us the opportunity to extend the usual meaning of the word “sum,” which

always applies to a finite number of terms being added together. By using limits, we

can make sense of expressions in which an infinite number of terms are added together.

Such expressions are called .infinite series

4s1 2 s21/3dnd}}

1 2 s21/3d



The first thing to understand about an infinite series is that it is not a true sum. There

are properties of real number addition that allow us to extend the definition of a 1 b to

sums like a 1 b 1 c 1 d 1 e 1 f, but not to “infinite sums.” For example, we can add

any finite number of 2’s together and get a real number, but if we add an infinite num-

ber of 2’s together we do not get a real number at all. Sums do not behave that way.

What makes series so interesting is that sometimes (as in Example 2) the sequence of

, all of which are true sums, approaches a finite limit S:

limn→`

^n

k51

ak 5 limn→`

sa1 1 a2 1 ? ? ? 1 and 5 S.

In this case we say that the series to S, and it makes sense to define S as the

. In sigma notation,

^∞

k51

ak 5 limn→`

^n

k51

ak 5 S.

If the limit of partial sums does not exist, then the series and has no sum.diverges

sum of the infinite series

converges

partial sums

EXAMPLE 3 Looking at Limits of Partial Sums

For each of the following series, find the first five terms in the sequence of partial

sums. Which of the series appear to converge?

(a) 0.1 1 0.01 1 0.001 1 0.0001 1 ? ? ?

(b) 10 1 20 1 30 1 40 1 ? ? ?

(c) 1 2 1 1 1 2 1 1 ? ? ?

SOLUTION

(a) The first five partial sums are h0.1, 0.11, 0.111, 0.1111, 0.11111j. These appear to

be approaching a limit of 0.1# 5 1/9, which would suggest that the series converges

to a sum of 1/9.

(b) The first five partial sums are h10, 30, 60, 100, 150j. These numbers increase

without bound and do not approach a limit. The series diverges and has no sum.

(c) The first five partial sums are h1, 0, 1, 0, 1j. These numbers oscillate and do not

approach a limit. The series diverges and has no sum.


You might have been tempted to “pair off” the terms in Example 3c to get an infi-

nite summation of 0’s (and hence a sum of 0), but you would be applying a rule

(namely the associative property of addition) that works on finite sums but not, in

general, on infinite series. The sequence of partial sums does not have a limit, so any

manipulation of the series in Example 3c that appears to result in a sum is actually

meaningless.

ALERT

Some students will expect a series with

r 5 21 to converge because the partial

sums do not go to infinity. Explain that the

partial sums alternate between two values,

so the series does not converge.


Convergence of Geometric Series

Determining the convergence or divergence of infinite series is an important part of a

calculus course, in which series are used to represent functions. Most of the conver-

gence tests are well beyond the scope of this course, but we are in a position to settle

the issue completely for geometric series.


Proof

If r 5 1, the series is a 1 a 1 a 1 ? ? ?, which is unbounded and hence diverges.

If r 5 21, the series is a 2 a 1 a 2 a 1 ? ? ?, which diverges. (See Example 3c.)

If r Þ 1, then by the Sum of a Finite Geometric Sequence Theorem, the nth partial sum

of the series is ^n

k51a • rk21 5 as1 2 rnd/s1 2 rd. The limit of the partial sums is

limn→`fas1 2 rnd/s1 2 rdg, which converges if and only if limn→`rn is a finite number.

But limn→`rn is 0 when )r ) , 1 and unbounded when )r ) . 1. Therefore, the sequence

of partial sums converges if and only if )r ) , 1, in which case the sum of the series is

limn→`

fas1 2 rnd/ s1 2 rdg 5 as1 2 0d/s1 2 rd 5 a/s1 2 rd.

TEACHING NOTE

The grapher allows students to visualize the

convergence of an infinite geometric series.

The ease of visual understanding was not

available before recent technological

advances.

EXAMPLE 4 Summing Infinite Geometric Series

Determine whether the series converges. If it converges, give the sum.

(a) ^`

k51

3s0.75dk21 (b) ^`

n50(2}

4

5} )

n

(c) ^`

n51( }

p

2} )

n

(d) 1 1 }1

2} 1 }

1

4} 1 }

1

8} 1 ? ? ?

SOLUTION

(a) Since )r ) 5 )0.75 ) , 1, the series converges. The first term is 3s0.75d0 5 3, so the

sum is a/s1 2 rd 5 3/s1 2 0.75d 5 12.

(b) Since ) r ) 5 )24/5 ) , 1, the series converges. The first term is s24/5d0 5 1, so

the sum is a/s1 2 rd 5 1/s1 2 s24/5dd 5 5/9.

(c) Since )r ) 5 )p/2) . 1, the series diverges.

(d) Since )r ) 5 )1/2 ) , 1, the series converges. The first term is 1, and so the sum is

a/s1 2 rd 5 1/s1 2 1/2d 5 2.


NOTES ON EXAMPLES

Example 4d can be easily demonstrated

with a blank sheet of paper. Tear the paper

in half and label one half “1/2.” Tear the

remaining half in half and label one part

“1/4.” Tear the remaining fourth in half and

label one part “1/8.” Continue this process

until the students are able to visualize that

by putting the fractional parts back together

again they will very nearly make one

“whole” piece of paper with an “infinitely

small” portion missing.

THEOREM Sum of an Infinite Geometric Series

The geometric series ^`

k51a • rk21 converges if and only if )r ) , 1. If it does con-

verge, the sum is a/s1 2 rd.

5144_Demana_Ch09pp699-790 1/16/06 11:54 AM Page 748


QUICK REVIEW 9.5 (For help, see Section 9.4.)

In Exercises 1–4, hanj is arithmetic. Use the given information to find

a10.

1. a1 5 4; d 5 2 22 2. a1 5 3; a2 5 1 215

3. a3 5 6; a8 5 21 27 4. a5 5 3; an11 5 an 1 5 for n $ 1 28

In Exercises 5–8, hanj is geometric. Use the given information to find

a10.

5. a1 5 1; a2 5 2 512 6. a4 5 1; a6 5 2 8

7. a7 5 5; r 5 22 240 8. a8 5 10; a12 5 40 20

9. Find the sum of the first 5 terms of the sequence hn2j. 55

10. Find the sum of the first 5 terms of the sequence

h2n 2 1j. 25

EXAMPLE 5 Converting a Repeating Decimal

to Fraction Form

Express 0.2w3w4w 5 0.234234234 . . . in fraction form.

SOLUTION We can write this number as a sum: 0.234 1 0.000234 1

0.000000234 1 ? ? ?.

This is a convergent infinite geometric series in which a 5 0.234 and r 5 0.001. The

sum is

}1 2

a

r} 5 }

1 2

0.2

0

3

.0

4

01} 5 }

0

0

.

.

2

9

3

9

4

9} 5 }

2

9

3

9

4

9} 5 }

1

2

1

6

1}.


SECTION 9.5 EXERCISES

In Exercises 1–6, write each sum using summation notation, assuming

the suggested pattern continues.

1. 27 2 1 1 5 1 11 1 ? ? ? 1 53

2. 2 1 5 1 8 1 11 1 ? ? ? 1 29

3. 1 1 4 1 9 1 ? ? ? 1 sn 1 1d2

4. 1 1 8 1 27 1 ? ? ? 1 sn 1 1d3

5. 6 2 12 1 24 2 48 1 ? ? ?

6. 5 2 15 1 45 2 135 1 ? ? ?

In Exercises 7–12, find the sum of the arithmetic sequence.

7. 27, 23, 1, 5, 9, 13 18

8. 28, 21, 6, 13, 20, 27 57

9. 1, 2, 3, 4, . . . , 80 3240

10. 2, 4, 6, 8, . . . , 70 1260

11. 117, 110, 103, . . . , 33 975

12. 111, 108, 105, . . . , 27 2001

In Exercises 13–16, find the sum of the geometric sequence.

13. 3, 6, 12, . . . , 12,288 24,573

14. 5, 15, 45, . . . , 98,415 147,620

15. 42, 7, }7

6}, . . . , 42 (}

1

6})

8

50.4(1 2 629) < 50.4

16. 42, 27, }7

6}, . . . , 42 (2 }

1

6})

9

36 2 628 < 36

In Exercises 17–22, find the sum of the first n terms of the sequence.

The sequence is either arithmetic or geometric.

17. 2, 5, 8, . . . ; n 5 10 155 18. 14, 8, 2, . . . ; n 5 9 290

19. 4, 22, 1, 2}1

2}, . . . ; n 5 12 }

8

3} (1 2 2212) < 2.666

20. 6, 23, }3

2}, 2}

3

4}, . . . ; n 5 11 4 (1 1 2211) < 4.002

21. 21, 11, 2121, . . . ; n 5 9 2196,495,641

22. 22, 24, 2288, . . . ; n 5 8 66,151,030



23. Find the first six partial sums of the following infinite series. If the

sums have a finite limit, write “convergent.” If not, write “diver-

gent.”

(a) 0.3 1 0.03 1 0.003 1 0.0003 1 ? ? ?

(b) 1 2 2 1 3 2 4 1 5 2 6 1 ? ? ?

24. Find the first six partial sums of the following infinite series. If the

sums have a finite limit, write “convergent.” If not, write “divergent.”

(a) 22 1 2 2 2 1 2 2 2 1 ? ? ? divergent

(b) 1 2 0.7 2 0.07 2 0.007 2 0.0007 2 ? ? ? convergent

In Exercises 25–30, determine whether the infinite geometric series

converges. If it does, find its sum.

25. 6 1 3 1 }3

2} 1 }

3

4} 1 ? ? ? 26. 4 1 }

4

3} 1 }

4

9} 1 }

2

4

7} 1 ? ? ?

27. }6

1

4} 1 }

3

1

2} 1 }

1

1

6} 1 }

1

8} 1 ? ? ? no

28. }4

1

8} 1 }

1

1

6} 1 }

1

3

6} 1 }

1

9

6} 1 ? ? ? no

29. ^`

j51

3 (}1

4})

j

yes; 1 30. ^`

n51

5 (}2

3})

n

yes; 10

In Exercises 31–34, express the rational number as a fraction of

integers.

31. 7.14141414 . . . 707/99 32. 5.93939393 . . . 196/33

33. 217.268268268 . . . 34. 212.876876876 . . .

35. Savings Account The table below shows the December bal-

ance in a fixed-rate compound savings account each year from

1996 to 2000.

37. Annuity Mr. O’Hara deposits $120 at the end of each month

into an account that pays 7% interest compounded monthly. After

10 years, the balance in the account, in dollars, is

12011 1 }0

1

.0

2

7}2

0

1 12011 1 }0

1

.0

2

7}2

1

1 ? ? ?

1 12011 1 }0

1

.0

2

7}2

119

.

(a) This is a geometric series. What is the first term? What

is r? 120; 1 1 0.07/12

(b) Use the formula for the sum of a finite geometric sequence to

find the balance. $20,770.18

38. Annuity Ms. Argentieri deposits $100 at the end of each month

into an account that pays 8% interest compounded monthly. After

10 years, the balance in the account, in dollars, is

10011 1 }0

1

.0

2

8}2

0

1 10011 1 }0

1

.0

2

8}2

1

1 ? ? ?

1 10011 1 }0

1

.0

2

8}2

119

.

(a) This is a geometric series. What is the first term? What

is r? 100; 1 1 0.08/12

(b) Use the formula for the sum of a finite geometric sequence to

find the balance. $18,294.60

39. Group Activity Follow the Bouncing Ball When “super-

balls” sprang upon the scene in the 1960s, kids across the United

States were amazed that these hard rubber balls could bounce to

90% of the height from which they were dropped. If a superball is

dropped from a height of 2 m, how far does it travel by the time it

hits the ground for the tenth time? [Hint: The ball goes down to

the first bounce, then up and down thereafter.] < 24.05 m

40. Group Activity End Behavior Does the graph of

f sxd 5 2 ( )have a horizontal asymptote to the right? How does this relate to

the convergence or divergence of the series

2 1 2.1 1 2.205 1 2.31525 1 ? ? ? ?

Standardized Test Questions41. True or False If all terms of a series are positive, the series

sums to a positive number. Justify your answer.

42. True or False If ^`

n51

an and ^`

n51

bn both diverge, then ^`

n51

(an 1 bn)

diverges. Justify your answer.

You should solve Exercises 43–46 without the use of a calculator.

43. Multiple Choice The series 321 1 322 1 323 1 ? ? ? 1

32n 1 ? ? ? A

(A) converges to 1/2 (B) converges to 1/3 (C) converges to 2/3

(D) converges to 3/2 (E) diverges

1 2 1.05x

}}1 2 1.05

Year 1996 1997 1998 1999 2000

Balance $20,000 $22,000 $24,200 $26,620 $29,282

(a) The balances form a geometric sequence. What is r? 1.1

(b) Write a formula for the balance in the account n years after

December 1996. 20,000(1.1)n

(c) Find the sum of the December balances from 1996 to 2006,

inclusive. $370,623.34

36. Savings Account The table below shows the December bal-

ance in a simple interest savings account each year from 1996 to

2000.

Year 1996 1997 1998 1999 2000

Balance $18,000 $20,016 $22,032 $24,048 $26,064

(a) The balances form an arithmetic sequence. What is d?

(b) Write a formula for the balance in the account n years after

December 1996. 18,000 1 2016n

(c) Find the sum of the December balances from 1996 to 2006,

inclusive. $308,880


44. Multiple Choice If ^`

n51

xn 5 4, then x 5 D

(A) 0.2. (B) 0.25. (C) 0.4. (D) 0.8. (E) 4.0.

45. Multiple Choice The sum of an infinite geometric series with

first term 3 and second term 0.75 is C

(A) 3.75. (B) 2.4. (C) 4. (D) 5. (E) 12.

46. Multiple Choice ^`

n50

412}5

3}2

n

5 E

(A) 26 (B) 2}5

2} (C) }

3

2} (D) 10 (E) divergent

Explorations47. Population Density The National Geographic Picture Atlas of

Our Fifty States (2001) groups the states into 10 regions. The two

largest groupings are the Heartland (Table 9.1) and the Southeast

(Table 9.2). Population and area data for the two regions are given

in the tables. The populations are official 2000 U.S. Census figures.

(a) What is the total population of each region?

(b) What is the total area of each region?

(c) What is the population density (in persons per square mile) of

each region?

(d) Writing to Learn For the two regions, compute the popu-

lation density of each state. What is the average of the seven

state population densities for each region? Explain why these

answers differ from those found in part (c).

48. Finding a Pattern Write the finite series 21 1 2 1 7 1 14 1

23 1 ? ? ? 1 62 in summation notation.

Table 9.2 The Southeast

State Population Area (mi2)

Alabama 4,447,100 51,705

Arkansas 2,673,400 53,187

Florida 15,982,378 58,644

Georgia 8,186,453 58,910

Louisiana 4,468,976 47,751

Mississippi 2,844,658 47,689

S. Carolina 4,012,012 31,113

Table 9.1 The Heartland

State Population Area (mi2)

Iowa 2,926,324 56,275

Kansas 2,688,418 82,277

Minnesota 4,919,479 84,402

Missouri 5,595,211 69,697

Nebraska 1,711,283 77,355

North Dakota 642,200 70,703

South Dakota 754,844 77,116

Extending the Ideas49. Fibonacci Sequence and Series Complete the following

table, where Fn is the nth term of the Fibonacci sequence and Sn is

the nth partial sum of the Fibonacci series. Make a conjecture

based on the numerical evidence in the table.

Sn 5 ^n

k51

Fk

n Fn Sn Fn12 2 1

1 1 1 1

2 1 2 2

3 2 4 4

4 3 7 7

5 5 12 12

6 8 20 20

7 13 33 33

8 21 54 54

9 34 88 88

Conjecture: Sn 5 Fn12 2 1

50. Triangular Numbers Revisited Exercise 41 in Section 9.2

introduced triangular numbers as numbers that count objects

arranged in triangular arrays:


1 151063

In that exercise, you gave a geometric argument that the nth trian-

gular number was nsn 1 1d/2. Prove that formula algebraically

using the Sum of a Finite Arithmetic Sequence Theorem.

51. Square Numbers and Triangular Numbers Prove that the

sum of two consecutive triangular numbers is a square number;

that is, prove

Tn21 1 Tn 5 n2

for all positive integers n $ 2. Use both a geometric and an alge-

braic approach.

52. Harmonic Series Graph the sequence of partial sums of the

harmonic series:

1 1 }1

2} 1 }

1

3} 1 }

1

4} 1 ? ? ? 1 }

1

n} 1 ? ? ? .

Overlay on it the graph of f sxd 5 ln x. The resulting picture

should support the claim that

1 1 }1

2} 1 }

1

3} 1 }

1

4} 1 ? ? ? 1 }

1

n} $ ln n,

for all positive integers n. Make a table of values to further sup-

port this claim. Explain why the claim implies that the harmonic

series must diverge.


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5144 Demana Ch09pp699-790