Answer
$-21846$
Work Step by Step
Here, $a_1=-2, r=-2, n=15$
Now, we have
$S_n=\dfrac{a_1(1-r^n)}{1-r}$
Now,
$S_{15}=\dfrac{(-2)(1-(-2)^{15})}{1-(-2)}$
or, $=-21846$
Intermediate Algebra for College Students (7th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13417-894-7
ISBN 13:
978-0-13417-894-3
Chapter 11 - Test - Page 871: 9
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