Answer
See below.
Work Step by Step
$\sum_{i=1}^{4} (-1)^{k+1}\frac{k+1}{k^2}=(-1)^{1+1}\frac{1+1}{1^2}+(-1)^{2+1}\frac{2+1}{2^2}+(-1)^{3+1}\frac{3+1}{3^2}=(1)2+(-1)0.75+(1)\frac{4}{9}=\frac{72}{36}-\frac{27}{36}+\frac{16}{36}=\frac{61}{36}$
College Algebra (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0321979478
ISBN 13:
978-0-32197-947-6
Chapter 9 - Test - Page 680: 3
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