Answer
$\frac{61}{36}$
Work Step by Step
$\sum_{k=1}^{3} (-1)^{k+1}\frac{k+1}{k^2}\\=(-1)^{1+1}\frac{1+1}{1^2}+(-1)^{2+1}\frac{2+1}{2^2}+(-1)^{3+1}\frac{3+1}{3^2}\\=(-1)^{2}\frac{2}{1}+(-1)^{3}\frac{3}{4}+(-1)^{4}\frac{4}{9}\\=\frac{61}{36}$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 12 - Sequences; Induction; the Binomial Theorem - Chapter Review - Chapter Test - Page 840: 3
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