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B. Find the nth Term of a Geometric Sequence

B. Find the nth Term of a Geometric Sequence

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continue applying additional factors of r for each successive term.

a1 ϭ a1r0

6 ϭ 3 # 21

a2 ϭ a1r1

12 ϭ 3 # 22

a3 ϭ a1r2

24 ϭ 3 # 23

a4 ϭ a1r3

48 ϭ 3 # 24

a5 ϭ a1r4

current term

initial

term

S

3 ϭ 3 # 20

S

784

exponent on common ratio

From this pattern, we note the exponent on r is always 1 less than the subscript of

the current term: 5 Ϫ 1 ϭ 4, which leads us to the formula for the nth term of a geometric sequence.

The n th Term of a Geometric Sequence

The nth term of a geometric sequence is given by

an ϭ a1rnϪ1

where r is the common ratio.

EXAMPLE 3

Finding a Specific Term in a Sequence

Identify the common ratio r, and use it to write the expression for the nth term.

Then find the 10th term of the sequence: 3, Ϫ6, 12, Ϫ24, p .

Solution

By inspection we note that a1 ϭ 3 and r ϭ Ϫ2. This gives

an ϭ a1rnϪ1

ϭ 31Ϫ22 nϪ1

n th term formula

substitute 3 for a1 and Ϫ2 for r

To find the 10th term we substitute n ϭ 10:

a10 ϭ 31Ϫ22 10Ϫ1

ϭ 31Ϫ22 9 ϭ Ϫ1536

substitute 10 for n

simplify

Now try Exercises 33 through 46

EXAMPLE 4

Determining the Number of Terms in a Geometric Sequence

1

.

Find the number of terms in the geometric sequence 4, 2, 1, p , 64

Solution

Observing that a1 ϭ 4 and r ϭ 12, we have

an ϭ a1 rnϪ1

1 nϪ1

ϭ 4a b

2

n th term formula

substitute 4 for a1 and

1

for r

2

Although we don’t know the number of terms in the sequence, we do know the last

1

1

. Substituting an ϭ 64

or nth term is 64

gives

1

1 nϪ1

ϭ 4a b

64

2

1

1 nϪ1

ϭa b

256

2

substitute

1

for an

64

1

divide by 4 amultiply by b

4

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Section 9.3 Geometric Sequences

785

From our work in Chapter 5, we attempt to write both sides as exponentials with a

like base, or apply logarithms. Since 256 ϭ 28, we equate bases.

1 8

1 nϪ1

a b ϭa b

2

2

S8 ϭ n Ϫ 1

9ϭn

write

1

1 8

as a b

256

2

like bases imply exponents must be equal

solve for n

This shows there are nine terms in the sequence.

Now try Exercises 47 through 58

Note that in both Examples 3 and 4, the nth term had the form of an exponential

1 nϪ1

equation 1y ϭ a # bn 2 after simplifying: an ϭ 31Ϫ22 nϪ1 and an ϭ 4a b . This is in

2

fact a characteristic of geometric sequences, with the common ratio r corresponding to

the base b. This means the graph of a geometric sequence will always be a set of discrete points that lie on an exponential graph (see Worthy of Note). After entering

1 nϪ1

(from Example 4), Figure 9.35 shows the table of values for this

u1n2 ϭ 4a b

2

sequence, with the graph in Figure 9.36.

As before, we can see the graph of the sequence more distinctly by entering the

natural numbers 1 through 8 in L1, then defining L2 as u(L1). The resulting graph is

shown in Figure 9.37.

WORTHY OF NOTE

The sequence from Example 3 is an

alternating sequence, and the

exponential characteristics of its

graph can be seen by taking a look

at the graph of an ϭ |31Ϫ22 nϪ1 | .

Figure 9.35

Figure 9.36

Figure 9.37

5

5

10

0

Ϫ1.5

10

0

Ϫ1.5

One additional advantage of using a list is that we can easily verify whether or not

a common ratio r exists. We do this by duplicating L2 in L3 (define L3 ϭ L2), then

deleting the first entry of L3 and the last entry of L2 (to keep the same number of terms

akϩ1

in each list). See Figure 9.38. This enables us to find the ratio

for the entire list

ak

by defining L4 as the ratio L3/L2 (which automatically computes the ratio of successive terms). See Figures 9.39 and 9.40.

Figure 9.38

Figure 9.39

Figure 9.40

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CHAPTER 9 Additional Topics in Algebra

EXAMPLE 5

Graphing Geometric Sequences

Enter the natural numbers 1 through 6 in L1, and the terms of the sequence

0.25, 1.25, 6.25, 31.25, 156.25, 781.25 in L2. Then determine if the sequence is

geometric by

a. Graphing the related points to see if they appear to form along an exponential

graph.

b. Finding the successive ratios between terms.

c. If the sequence is geometric, find the nth term and graph the sequence.

Solution

a. The plotted points are shown in Figure 9.41 and appear to lie along an

exponential graph.

Figure 9.41

Figure 9.42

1000

8

0

Ϫ250

b. Using the approach described prior to

Example 5, we find there is a common

ratio of r ϭ 5 (Figure 9.42).

c. With a1 ϭ 0.25 and r ϭ 5, the nth term

must be

an ϭ a1rnϪ1

ϭ 0.25152 nϪ1

Figure 9.43

1000

n th term formula

8

0

substitute for a1 and r

The nth term for this sequence is

an ϭ 0.25152 nϪ1. The graph is shown in

Figure 9.43.

Ϫ250

Now try Exercises 59 through 62

If the term a1 is unknown but a term ak is given, the nth term can be written

an ϭ akrnϪk,

(the subscript on the term ak and the exponent on r sum to n).

EXAMPLE 6

Finding the First Term of a Geometric Sequence

Given a geometric sequence where a4 ϭ 0.075 and a7 ϭ 0.009375, find the

common ratio r and the value of a1.

Solution

Since a1 is not known, we express a7 as the product of a known term and the

appropriate number of common ratios: a7 ϭ a4r3 17 ϭ 4 ϩ 3, as required).

a7 ϭ a4 # r3

0.009375 ϭ 0.075r3

0.125 ϭ r3

r ϭ 0.5

a1 is unknown

substitute 0.009375 for a7 and 0.075 for a4

divide by 0.075

solve for r

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Section 9.3 Geometric Sequences

Having found r, we can now solve for a1

a7 ϭ a1r6

0.009375 ϭ a1 10.52 6

0.009375 ϭ a1 10.0156252

a1 ϭ 0.6

n th term formula

substitute 0.009375 for a7 and 0.5 for r

simplify

solve for a1

The first term is a1 ϭ 0.6 and the common ratio is r ϭ 0.5.

B. You’ve just seen how

we can find the nth term of a

geometric sequence

Now try Exercises 63 through 68

C. Find the n th Partial Sum of a Geometric Sequence

As with arithmetic series, applications of geometric series often involve computing sums

of consecutive terms. We can adapt the method for finding the sum of an arithmetic

sequence to develop a formula for adding the first n terms of a geometric sequence.

For the nth term an ϭ a1rnϪ1, we have Sn ϭ a1 ϩ a1r ϩ a1r2 ϩ a1r3 ϩ p ϩ

nϪ1

a1r . If we multiply Sn by Ϫr then add the original series, the “interior terms” sum

to zero.

Ϫ rSn ϭ Ϫa1r ϩ 1Ϫa1r2 2 ϩ 1Ϫa1r 3 2 ϩ p ϩ 1Ϫa1rnϪ1 2 ϩ 1Ϫa1rn 2

------S --S --S

S a rnϪ1

ϩ Sn ϭ a1 ϩ a1r ϩS a1r 2 ϩ p ϩ a1rnϪ2 ϩ

1

Sn Ϫ rSn ϭ

a1 ϩ

0

ϩ

0

ϩ 0 ϩ

0

ϩ

0

ϩ 1Ϫa1rn 2

We then have Sn Ϫ rSn ϭ a1 Ϫ a1rn, and can now solve for Sn:

Sn 11 Ϫ r2 ϭ a1 Ϫ a1rn

factor out Sn

a1 Ϫ a1r

solve for Sn (divide by 1 Ϫ r )

1Ϫr

The result is a formula for the nth partial sum of a geometric sequence.

n

Sn ϭ

The n th Partial Sum of a Geometric Sequence

Given a geometric sequence with first term a1 and common ratio r, the nth partial

sum (the sum of the first n terms) is

a1 11 Ϫ rn 2

a1 Ϫ a1rn

Sn ϭ

ϭ

,r 1

1Ϫr

1Ϫr

In words: The sum of the first n terms of a geometric sequence is the difference of

the first and 1n ϩ 12st term, divided by 1 minus the common ratio.

EXAMPLE 7

Solution

Computing a Partial Sum

3i (the first nine powers

Use the preceding summation formula to find the sum:

iϭ1

of 3). Verify the result on a graphing calculator.

The initial terms of this series are 3 ϩ 9 ϩ 27 ϩ p , and we note a1 ϭ 3, r ϭ 3,

and n ϭ 9. We could find the first nine terms and add, but using the partial sum

formula is much faster and gives

Sn ϭ

S9 ϭ

a1 11 Ϫ rn 2

1Ϫr

311 Ϫ 39 2

1Ϫ3

9

͚

sum formula

substitute 3 for a1, 9 for n, and 3 for r

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ϭ

31Ϫ19,6822

Ϫ2

ϭ 29,523

simplify

result

To verify, we enter u1n2 ϭ 3 on the Y= screen,

and find the sum of the first nine terms of the

sequence on the home screen as before. See figure.

n

C. You’ve just seen how

we can find the nth partial sum

of a geometric sequence

Now try Exercises 69 through 92

D. The Sum of an Infinite Geometric Series

To this point we’ve considered only partial sums of a geometric series. While it is

impossible to add an infinite number of these terms, some of these “infinite sums” appear

to have what is called a limiting value. The sum appears to get ever closer to this value

but never exceeds it—much like the asymptotic behavior of some graphs. We will define

the sum of this infinite geometric series to be this limiting value, if it exists. Consider

the illustration in Figure 9.44, where a standard sheet of typing paper is cut in half. One

of the halves is again cut in half and the process is continued indefinitely, as shown.

1 1 p

, 32, with a1 ϭ 12 and r ϭ 12.

Notice the “halves” create an infinite sequence 21, 14, 18, 16

1

1

1

1

1

The corresponding infinite series is 2 ϩ 4 ϩ 8 ϩ 16 ϩ 32 ϩ p ϩ 21n ϩ p .

Figure 9.44

1

2

1

8

1

2

1

4

1

4

1

8

1

16

1

16

1

32

1

32

1

64

1

64

and so on

Figure 9.45

1

64

1

16

1

32

1

4

1

8

1

2

WORTHY OF NOTE

The formula for the sum of an

infinite geometric series can

also be derived by noting that

Sq ϭ a1 ϩ a1r ϩ a1r2 ϩ

a1r3 ϩ p can be rewritten as

Sq ϭ a1 ϩ r1a1 ϩ a1r ϩ

a1r2 ϩ a1r3 ϩ p 2 ϭ a1 ϩ rSq.

Sq Ϫ rSq ϭ a1

Sq 11 Ϫ r2 ϭ a1

a1

.

Sq ϭ

1Ϫr

If we arrange one of the halves from each stage as shown in Figure 9.45, we would be

rebuilding the original sheet of paper. As we add more and more of these halves together,

we get closer and closer to the size of the original sheet. We gain an intuitive sense that this

series must add to 1, because the pieces of the original sheet of paper must add to 1 whole

sheet. To explore this idea further, consider what happens to 1 12 2 n as n becomes large.

1 4

n ϭ 4: a b ϭ 0.0625

2

1 8

n ϭ 8: a b Ϸ 0.004

2

1 12

n ϭ 12: a b Ϸ 0.0002

2

Further exploration with a calculator seems to support the idea that as

n S q, 1 12 2 n S 0, although a definitive proof is left for a future course. In fact, it can

be shown that for any ͿrͿ 6 1, rn becomes very close to zero as n becomes large.

a1 Ϫ a1rn

a1

a1rn

n

In symbols: as n S q, r S 0. For Sn ϭ

ϭ

Ϫ

, note that if

1Ϫr

1Ϫr

1Ϫr

ͿrͿ 6 1 and “we sum an infinite number of terms,” the second term becomes zero, leaving

a1

.

only the first term. In other words, the limiting value (represented by Sq) is Sq ϭ

1Ϫr

Infinite Geometric Series

Given a geometric sequence with first term a1 and 0 r 0 6 1, the sum of the related

infinite series is given by

a1

Sq ϭ

;r 1

1Ϫr

If ͿrͿ 7 1, no finite sum exists.

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