Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - Chapter Review Exercises - Page 591: 30

Answer

$$ \frac{4}{33} $$

Work Step by Step

Since \begin{align*} 0.121212 &= 0.12+0.0012+0.000012\\ &=0.12(1)+0.12(10^{-2})+ 0.12(10^{-4})+\cdots \\ &=0.12[1+\frac{1}{10^2}+\frac{1}{10^4}+\cdots ] \end{align*} Since $$1+\frac{1}{10^2}+\frac{1}{10^4}+\cdots $$ which is a geometric series and has the sum $$\frac{1}{1-\frac{1}{100}}= \frac{100}{99} $$ Hence $$0.12121212= 0.12 \frac{100}{99}= \frac{12}{99} =\frac{4}{33}$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.