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: 50

Answer

$=\dfrac {124}{99}$

Work Step by Step

$1.\overline {25}=1.2525\ldots =1+\dfrac {25}{100}+\dfrac {25}{100^{2}}+\dfrac {25}{1003}+\ldots =1+\dfrac {a_{1}}{1-r}=1+\dfrac {\dfrac {25}{100}}{1-\dfrac {1}{100}}=1+\dfrac {25}{99}=\dfrac {124}{99}$
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