College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Summary, Review, and Test - Review Exercises - Page 789: 45



Work Step by Step

We can see that the ratio of subsequent terms is $\frac{1}{4}$, thus $r=\frac{1}{4}$. The sum of the first $n$ terms can be obtained by multiplying the first term($a_1$) by the difference of $1$ and the common ratio ($r$) on the power of $n$ divided by the difference of $1$ and the common ratio ($r$). So $S_n=\frac{a_1(1-r^n)}{1-r}$. Hence here $S_{5}=\frac{2(1-(\frac{1}{4})^{5})}{1-\frac{1}{4}}=\frac{8}{3}$.
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