Calculus 8th Edition

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

Appendix E - Sigma Notation - E Exercises - Page A38: 34


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

Work Step by Step

Find the value of the sum $\sum\limits_{i =1}^{n}i(i+1)(i+2)$ After expanding the terms , we have $\sum\limits_{i =1}^{n}i(i+1)(i+2)=\sum\limits_{i =1}^{n}(i^{3}+3i^{2}+2i)$ $=\sum\limits_{i =1}^{n}i^{3}+3\sum\limits_{i =1}^{n}i^{2}+2\sum\limits_{i =1}^{n}i$ $=[\frac{n(n+1)}{2}]^{2}+3[\frac{n(n+1)(2n+1)}{6}]+2[\frac{n(n+1)}{2}]$ $=\frac{1}{4}n^{2}(n+1)^{2}+\frac{1}{2}n(n+1)(2n+1)+n(n+1)$ $=\frac{1}{4}n(n+1)[n^{2}+n+4n+2+4]$ $=\frac{1}{4}n(n+1)[n^{2}+5n+6]$ Hence, $\sum\limits_{i =1}^{n}i(i+1)(i+2)=\frac{1}{4}n(n+1)(n+2)(n+3)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.