Answer
$2^2+3^2+...+(n+1)^2$
Work Step by Step
$\sum_{k=1}^{n} (k+1)^2=(1+1)^2+(2+1)^2+...+(n+1)^2=2^2+3^2+...+(n+1)^2$
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 - 12.1 Sequences - 12.1 Assess Your Understanding - Page 807: 52
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