Chapter 7: Q. 38 (page 615)

Evaluate the finite sums.
$\sum_{k = 1}^{50} \frac{1}{3^{k}}$

Short Answer

Expert verified

The sum of the series $\sum_{k = 1}^{50} \frac{1}{3^{k}}$ is, $\frac{1 - {(\frac{1}{3})}^{50}}{2}$ .

Step by step solution

Step 1. Given Information

$\sum_{k = 1}^{50} \frac{1}{3^{k}}$ .

Step 2. Let us expand the given series.

$\sum_{k = 1}^{50} \frac{1}{3^{k}} = \frac{1}{3} + \frac{1}{3^{2}} + . . . . + \frac{1}{3^{50}}$

The first term in the series $a = \frac{1}{3}$

The common ratio is, $\frac{1}{3}$ with the number of terms, $n = 50$ .

Step 3. The finite sum of the geometric series with ratio less than 1 is given by,

$S_{n} = \frac{a (1 - r^{n})}{1 - r}$

Substitute the known values in the formula.

$S_{50} = \frac{\frac{1}{3} (1 - {(\frac{1}{3})}^{50})}{1 - \frac{1}{3}} = \frac{\frac{1}{3} - {(\frac{1}{3})}^{51}}{\frac{2}{3}} = \frac{1 - {(\frac{1}{3})}^{50}}{2}$

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Evaluate the finite sums.
$\sum_{k = 1}^{50} \frac{1}{3^{k}}$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Let us expand the given series.

Step 3. The finite sum of the geometric series with ratio less than 1 is given by,

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Evaluate the finite sums.∑k=150 13k

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Let us expand the given series.

Step 3. The finite sum of the geometric series with ratio less than 1 is given by,

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

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Study anywhere. Anytime. Across all devices.

Evaluate the finite sums.
$\sum_{k = 1}^{50} \frac{1}{3^{k}}$