Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Appendix E - Sigma Notation - E Exercises - Page A38: 35


$\sum\limits_{i =1}^{n}(i^{3}-i-2)=\frac{1}{4}n(n^{3}+2n^{2}-n-10)$

Work Step by Step

Find the value of the sum $\sum\limits_{i =1}^{n}(i^{3}-i-2)$ After expanding the terms , we have $\sum\limits_{i =1}^{n}(i^{3}-i-2)=\sum\limits_{i =1}^{n}i^{3}-\sum\limits_{i =1}^{n}i-2\sum\limits_{i =1}^{n}1$ $=[\frac{n(n+1)}{2}]^{2}-[\frac{n(n+1)}{2}]-2n$ $=\frac{1}{4}n^{2}(n+1)^{2}+\frac{1}{2}n(n+1)-2n$ $=\frac{1}{4}n[n(n+1)^{2}-2(n+1)-8]$ Hence, $\sum\limits_{i =1}^{n}(i^{3}-i-2)=\frac{1}{4}n(n^{3}+2n^{2}-n-10)$
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