University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 498: 63



Work Step by Step

The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ Since, we have $\sum_{n =1}^{ \infty}\dfrac{2^n+3^n}{4^n}=\sum_{n =1}^{ \infty} (\dfrac{1}{2})^n+\sum_{n =0}^{ \infty} (\dfrac{3}{4})^n$ Both series are convergent geometric series with first term, $a=\dfrac{1}{2},\dfrac{3}{4}$ and common ratio $r =\dfrac{1}{2},\dfrac{3}{4}$ Thus, $S=\dfrac{1/2}{1-\dfrac{1}{2}}+\dfrac{3/4}{1-\dfrac{3}{4}}=1+3=4$
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.