Answered: 8 k=1 | bartleby

Algebra: Structure And Method, Book 1

(REV)00th Edition

ISBN:9780395977224

Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Chapter4: Polynomials

Section4.3: Multiplying Monomials

Problem 2WE

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Question

Evaluate the geometric series or state that it diverges.

Expert Solution

Step 1

Given: $S = \sum_{k = 1}^{\infty} 3 {(- \frac{1}{8})}^{3 k}$

for evaluating given summation, first we expand it

so,

$\begin{array}{rcl} S & = & \sum_{k = 1}^{\infty} 3 {(- \frac{1}{8})}^{3 k} \\ = & 3 \sum_{k = 1}^{\infty} {(- \frac{1}{8})}^{3 k} \\ = & 3 [{(- \frac{1}{8})}^{3} + {(- \frac{1}{8})}^{6} + {(- \frac{1}{8})}^{9} + . . . . . . . . . . . . . . . . . . . .] . . (1) \end{array}$

now, we can see that equation $(1)$ is infinite geometric series

Step 2

we know that sum $(S)$ of infinite geometric series is given by

$S = \frac{a}{1 - r} . . (2)$

where; a is first term and r is common ratio

here, $a = {(- \frac{1}{8})}^{3}, r = {(- \frac{1}{8})}^{3}$

so, given summation will be

$S = 3 [\frac{{(- \frac{1}{8})}^{3}}{1 - {(- \frac{1}{8})}^{3}}] = 3 [\frac{- \frac{1}{8^{3}}}{1 + \frac{1}{8^{3}}}] = 3 [\frac{- \frac{1}{8^{3}}}{\frac{8^{3} + 1}{8^{3}}}] = 3 [- \frac{1}{8^{3} + 1}] = - \frac{3}{8^{3} + 1} = - \frac{3}{512 + 1} = - \frac{3}{513}$

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