University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 497: 14



Work Step by Step

We have the geometric series: $2+\dfrac{4}{5}+\dfrac{8}{25}+\dfrac{16}{25}+...$ or, $2(1+\dfrac{2}{5}+\dfrac{4}{25}+\dfrac{8}{25}+...)$ or, $2(1+\dfrac{2}{5}+(\dfrac{2}{5})^2+(\dfrac{2}{5})^3+...)$ Here, $a=1, r=\dfrac{2}{5}$ The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}=2(\dfrac{1}{1-\dfrac{2}{5}})=\dfrac{10}{3}$
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