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: 11

Answer

$\dfrac{23}{2}$

Work Step by Step

Consider two geometric series: $(5+1)+(\dfrac{5}{2}+\dfrac{1}{3})+(\dfrac{5}{4}+\dfrac{1}{9})+(\dfrac{5}{8}+\dfrac{1}{27})+...$ Here, $a=5, r=\dfrac{1}{2}$ and $a=1, r=\dfrac{1}{3}$ Formula to find the sum of a geometric series is $S=S_1+S_2=\dfrac{a}{1-r}+\dfrac{a}{1-r}$ or, $S=\dfrac{5}{1-\dfrac{1}{2}}+\dfrac{1}{1-\dfrac{1}{3}}$ or, $S=10+\dfrac{3}{2}=\dfrac{23}{2}$
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