Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.3 Infinite Series - 8.3 Exercises - Page 623: 54


$=\dfrac {50771}{9900}$

Work Step by Step

$5.12\overline {83}=5.128383\ldots =\dfrac {512}{100}+\dfrac {1}{100}\left( \dfrac {83}{100}+\dfrac {83}{1002}+\dfrac {83}{1003}...\right) =\dfrac {512}{100}+\dfrac {1}{100}\left( \dfrac {a_{1}}{1-r}\right) =\dfrac {512}{100}+\dfrac {1}{100}\dfrac {\dfrac {83}{100}}{1-\dfrac {1}{100}}=\dfrac {1}{100}\left( 512+\dfrac {83}{99}\right) =\dfrac {50771}{9900}$
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