Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Review Exercises - Page 870: 42

Answer

$\dfrac{341}{128}$

Work Step by Step

Since, we have $S_n=\dfrac{a_1(1-r^{n})}{1-r}$ Here, $n=5,a_1=2,r=\dfrac{1}{4}$ Thus, $S_{5}=\dfrac{2[1-(\dfrac{1}{4})^{5}]}{1-(\dfrac{1}{4})}=\dfrac{8}{3}(\dfrac{1023}{1024})=\dfrac{341}{128}$
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