Algebra and Trigonometry 10th Edition

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


$\displaystyle \sum_{k=2}^{5}(k+1)^2(k-3)=88$

Work Step by Step

$\displaystyle \sum_{k=2}^{5}(k+1)^2(k-3)=(2+1)^2(2-3)+(3+1)^2(3-3)+(4+1)^2(4-3)+(5+1)^2(5-3)=-9+0+25+72=88$
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