Answer
$\frac{(k+1)(k+2)(2k+3)}{6}$
Work Step by Step
$\frac{k(k+1)(2k+1)}{6}+(k+1)^2=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}=\frac{(k+1)(k(2k+1)+6(k+1))}{6}=\frac{(k+1)(2k^2+k+6k+6)}{6}=\frac{(k+1)(k+2)(2k+3)}{6}$
College Algebra (6th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-32178-228-3
ISBN 13:
978-0-32178-228-1
Chapter 8 - Sequences, Induction, and Probability - Exercise Set 8.3 - Page 743: 114
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