Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 11 - 11.3 - Geometric Sequences and Series - 11.3 Exercises - Page 796: 66


$15-3+\frac{3}{5}-...-\frac{3}{625}=\displaystyle \sum_{n=1}^{6}15(-\frac{1}{5})^{n-1}$

Work Step by Step

$-\frac{3}{625}=-\frac{3}{625}\frac{15}{15}=15(-\frac{1}{3125})=15(-\frac{1}{5})^5$ $15-3+\frac{3}{5}-...-\frac{3}{625}=15(-\frac{1}{5})^0+15(-\frac{1}{5})^1+15(-\frac{1}{5})^2+...+15(-\frac{1}{5})^5=\displaystyle \sum_{n=1}^{6}15(-\frac{1}{5})^{n-1}$
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.