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: 20

Answer

$\dfrac{26}{111}$

Work Step by Step

Here, $0.234234234= \dfrac{234}{1000}+\dfrac{234}{(1000)^2}+\dfrac{234}{(1000)^3}+....=\dfrac{234}{1000}(1+ \dfrac{1}{1000}+ \dfrac{1}{(1000)^2}+..)$ Formula to find the sum of a geometric series $1+ \dfrac{1}{1000}+ \dfrac{1}{(1000)^2}+..$ is $S=\dfrac{a}{1-r}$ Here, $a=1, r=\dfrac{1}{1000}=0.001$; Now, $S=\dfrac{234}{1000}(\dfrac{1}{1-0.001})=\dfrac{26}{111}$
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