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: 49


$=\dfrac {4}{33}$

Work Step by Step

$0.\overline {12}=0.121212...=\dfrac {12}{100}+\dfrac {12}{100^{2}}+\dfrac {12}{100^{3}}+...=\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {12}{100}}{1-\dfrac {1}{100}}=\dfrac {\dfrac {12}{100}}{\dfrac {99}{100}}=\dfrac {12}{99}=\dfrac {4}{33}$
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