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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 12: Q 63. (page 825)

In Problems 51鈥�66, determine whether each infinite geometric series converges or diverges. If it converges, find its sum.
${鈭�}_{k = 1}^{鈭�} 6 {(- \frac{2}{3})}^{k - 1}$

Short Answer

Expert verified

The given infinite geometric series ${鈭�}_{k = 1}^{鈭�} 6 {(- \frac{2}{3})}^{k - 1}$ is convergent and its sum is $\frac{18}{5}$ .

Step by step solution

01

Step 1. Write the given information.

The given geometric series is:

${鈭�}_{k = 1}^{鈭�} 6 {(- \frac{2}{3})}^{k - 1}$

02

Determine the first term and common ratio.

The first term is $(a_{1} = 6)$

The common ratio is the ratio of successive terms:

$r = \frac{- 2}{3}$

03

Step 3. Determine whether the infinite geometric series converges or diverges.

If $|r| < 1$ then the infinite geometric series ${鈭�}_{k = 1}^{鈭�} a_{1} r^{k - 1}$ converges.

As we can see that $|- \frac{2}{3}| = \frac{2}{3} and \frac{2}{3} < 1$ , therefore the given series is convergent.

04

Step 4. Find the sum of the series.

Use the formula to find the sum of the given geometric series ${鈭�}_{k = 1}^{鈭�} a_{1} r^{k - 1} = \frac{a_{1}}{1 - r}$ ,

${鈭�}_{k = 1}^{鈭�} 6 {(- \frac{2}{3})}^{k - 1} = \frac{6}{(1 - (\frac{- 2}{3}))} = \frac{6}{\frac{5}{3}} = \frac{18}{5}$

Therefore, the sum of the infinite geometric series is $\frac{18}{5}$ .

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Most popular questions from this chapter

In Problems 71-82, find the sum of each sequence.

${鈭� k^{3}}_{k = 4}^{24}$

In Problems 71-82, find the sum of each sequence.

${鈭� (5}_{k = 1}^{20} k + 3)$

In Problems 37鈥�50, a sequence is defined recursively. Write down the first five terms.

$a_{1} = 3; a_{n} = 4 - a_{n - 1}$

In Problems 29鈥�36, the given pattern continues. Write down the nth term of a sequence $\{a_{n}\}$ suggested by the pattern.

$1, \frac{1}{2}, 3, \frac{1}{4}, 5, \frac{1}{6}, 7, \frac{1}{8}, . . .$

In Problems 37鈥�50, a sequence is defined recursively. Write down the first five terms.

$a_{1} = A; a_{n} = r a_{n - 1}, r 鈮� 0$

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