Calculus 8th Edition

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

Appendix E - Sigma Notation - E Exercises: 33

Answer

$\sum\limits_{i =1}^{n}(i+1)(i+2)=\frac{1}{3}n(n^{2}+6n+11)$

Work Step by Step

Find the value of the sum $\sum\limits_{i =1}^{n}(i+1)(i+2)$ After expanding the terms , we have $\sum\limits_{i =1}^{n}(i+1)(i+2)=\sum\limits_{i =1}^{n}(i^{2}+3i+2)$ $=\sum\limits_{i =1}^{n}i^{2}+3\sum\limits_{i =1}^{n}i+\sum\limits_{i =1}^{n}2$ $=[\frac{n(n+1)(2n+1)}{6}]+3[\frac{n(n+1)}{2}]+2n$ $=\frac{1}{3}n^{3}+2n^{2}+\frac{11}{3}n$ Hence, $\sum\limits_{i =1}^{n}(i+1)(i+2)=\frac{1}{3}n(n^{2}+6n+11)$
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