Answer
$P_{k+1}=\frac{1}{3}(k+1)(2k+3)$
Work Step by Step
$P_k=\frac{1}{3}k(2k+1)$
$P_{k+1}=\frac{1}{3}(k+1)[2(k+1)+1]=\frac{1}{3}(k+1)(2k+3)$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 11 - 11.4 - Mathematical Induction - 11.4 Exercises - Page 806: 8
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