Answer
104975
Work Step by Step
$\sum_{i=1}^{n}(i^3-2i)=\sum_{i=1}^ni^3-2\sum_{i=1}^ni$
$=\frac{n^2(n+1)^2}{4}-2\times\frac{n(n+1)}{2}=\frac{n^2(n+1)^2}{4}-n(n+1)$
Therefore,
$\sum_{i=1}^{25}(i^3-2i)=\frac{25^2(26)^2}{4}-25(26)=104975$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 4 - Integration - 4.2 Exercises - Page 263: 20
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