Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX E - Sigma Notation - E Exercises: 35

Answer

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

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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