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Chapter 7: Q. 16 (page 656)

Geometric series: For each of the series that follow, find the sum or explain why the series diverges.
${鈭�}_{k = 0}^{鈭�} \frac{5^{k + 3}}{2^{3 k + 1}}$

Short Answer

Expert verified

The sum of the series is $\frac{500}{3}$ .

Step by step solution

01

Step 1. Given Information

The given series is ${鈭�}_{k = 0}^{鈭�} \frac{5^{k + 3}}{2^{3 k + 1}}$ .

02

Step 2. Determine if a geometric series

Find the ratio of the consecutive terms.

$\begin{array}{rcl} \frac{\frac{5^{k + 3 + 1}}{2^{3 (k + 1) + 1}}}{\frac{5^{k + 3}}{2^{3 (k) + 1}}} & = & \frac{5^{k + 3 + 1 - k - 3}}{2^{3 (k + 1) + 1 - 3 k - 1}} \\ = & \frac{5}{2^{3}} \\ = & \frac{5}{8} \end{array}$

Since the ratio of the consecutive terms is same, the given series is a geometric series with a common ratio $r = \begin{array}{rcl} \frac{5}{8} \end{array}$ .

03

Step 3. Determine the convergence

Determine the first term and the common ratio.

$\begin{array}{rcl} c & = & \frac{5^{0 + 3}}{2^{3 (0) + 1}} \\ = & \frac{5^{3}}{2} \\ = & \frac{125}{2} \\ r & = & \frac{5}{8} \end{array}$

If $|r| < 1$ , the series converges to $\frac{c}{1 - r}$ .

$\begin{array}{rcl} \frac{c}{1 - r} & = & \frac{\frac{125}{2}}{1 - \frac{5}{8}} \\ = & \frac{\frac{125}{2}}{\frac{3}{8}} \\ = & \frac{500}{3} \end{array}$

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Most popular questions from this chapter

Determine whether the series ${鈭�}_{k = 0}^{鈭�} \frac{{(- 3)}^{k + 1}}{4^{k - 2}}$ converges or diverges. Give the sum of the convergent series.

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