Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.2 - Infinite Series - Exercises 10.2 - Page 579: 14

Answer

$\dfrac{10}{3}$

Work Step by Step

Given: $2+\dfrac{4}{5}+\dfrac{8}{25}+\dfrac{16}{25}+...$ The series can be re-written as: $2(1+\dfrac{2}{5}+\dfrac{4}{25}+\dfrac{8}{25}+...)$ or, $2(1+\dfrac{2}{5}+(\dfrac{2}{5})^2+(\dfrac{2}{5})^3+...)$ Here, $a=1, r=\dfrac{2}{5}$ Formula to find the sum of a geometric series is $S=\dfrac{a}{1-r}$ Thus, $S=2(\dfrac{1}{1-\dfrac{2}{5}})=\dfrac{10}{3}$
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