Answer
$S_{5}=31$
Work Step by Step
$\displaystyle \sum_{i=1}^{5}(2)^{i-1}=2^0+2^1+...+2^5$
$a_1=2^0=1$
$r=\frac{a_2}{a_1}=\frac{2^1}{2^0}=2$
$S_{5}=a_1(\frac{1-r^n}{1-r})=1(\frac{1-2^5}{1-2})=\frac{-31}{-1}=31$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 11 - Review Exercises - Page 841: 55
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