Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Review Exercises - Page 870: 38



Work Step by Step

Since, we have $S_n=\dfrac{a_1(1-r^{n})}{1-r}$ Here, $n=15,a_1=5,r=-3$ Thus, $S_{15}=\dfrac{5(1-(-3)^{15})}{1-(-3)}=17936135$
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