Answer
$\displaystyle \sum_{k=2}^{5}(k+1)^2(k-3)=88$
Work Step by Step
$\displaystyle \sum_{k=2}^{5}(k+1)^2(k-3)=(2+1)^2(2-3)+(3+1)^2(3-3)+(4+1)^2(4-3)+(5+1)^2(5-3)=-9+0+25+72=88$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 11 - 11.1 - Sequences and Series - 11.1 Exercises - Page 778: 71
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