Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 12 - Sequences and Series - 12.4 Infinite Series - 12.4 Exercises - Page 638: 22



Work Step by Step

The repeating decimal 0.18181818 . . . can be expressed as the infinite geometric series: $0.1818181818...=0.18+0.18(\frac{1}{100})+0.18(\frac{1}{100})^{2}+0.18(\frac{1}{100})^{3}...$ $a=0.18$ $r=\frac{1}{100}$ $\Sigma=\frac{0.18}{1-\frac{1}{100}}=\frac{2}{11}$ and $\frac{2}{11}$ is also $0.1818181818...$
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.