Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 756: 52

Answer

$\dfrac {46}{99}$

Work Step by Step

$0.\overline {46}=0.464646\ldots =\dfrac {46}{100}+\dfrac {46}{100^{2}}+\dfrac {46}{100^{3}}=\dfrac {a_{1}}{1-r};a_{1}=\dfrac {46}{100};r=\dfrac {1}{100}\Rightarrow 0.\overline {46}=\dfrac {\dfrac {46}{100}}{1-\dfrac {1}{100}}=\dfrac {46}{99}$
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