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: 33



Work Step by Step

The series $\sum_{n=-1}^{\infty} \frac{2^{n+3}}{3^{n}}$ can be rewritten as follows $$ \sum_{n=-1}^{\infty} \frac{2^{n+3}}{3^{n}}=\sum_{n=-1}^{\infty} 8\left(\frac{2}{3}\right)^{n}= 8\left(\frac{2}{3}\right)^{-1}+ \sum_{n=0}^{\infty} 8\left(\frac{2}{3}\right)^{n}. $$ Now, we have a geometric series $\sum_{n=0}^{\infty} 8\left(\frac{2}{3}\right)^{n}$ which has the sum $$\frac{c}{1-r}=\frac{8}{1-(2/3)}=24.$$ Hence, the sum of the series $\sum_{n=-1}^{\infty} \frac{2^{n+3}}{3^{n}}$ is $$12+24=36.$$
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.