Answered: E 12 n=1

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.5: The Binomial Theorem

Problem 14E

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Question

Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum.

(geometric series in picture)

Expert Solution

Step 1

Consider the following expression:

$\begin{array}{rcl} \sum_{n = 1}^{\infty} 12 {(\frac{1}{4})}^{n} & = & 12 \sum_{n = 1}^{\infty} {(\frac{1}{4})}^{n} \\ = & 12 [{(\frac{1}{4})}^{1} + {(\frac{1}{4})}^{2} + {(\frac{1}{4})}^{3} + {(\frac{1}{4})}^{4} + . . .] \\ = & 12 (\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + . . .) \end{array}$

Consider the following sequence:

$\frac{1}{4}, \frac{1}{16}, \frac{1}{64}, \frac{1}{256}, . . .$

$first term (a) = \frac{1}{4}$

$c o m m o n r a t i o (r) = \frac{\frac{1}{16}}{\frac{1}{4}} = \frac{\frac{1}{64}}{\frac{1}{16}} = \frac{1}{4}$

It is a geometric sequence.

Step 2

Infinite sum of geometric sequence:

$\begin{array}{rcl} S & = & \frac{a}{1 - r} \\ = & \frac{\frac{1}{4}}{1 - \frac{1}{4}} \\ = & \frac{\frac{1}{4}}{\frac{3}{4}} \\ = & \frac{1}{3} \end{array}$

$\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + . . . = \frac{1}{3}$

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