Calculus 8th Edition

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

Appendix E - Sigma Notation - E Exercises - Page A38: 31


The solution is $$\sum_{i=1}^n(i^2+3i+4)=\frac{1}{3}n\left(n^2+6n+17\right).$$

Work Step by Step

First, we will use the fact that the sum passes through terms individually, as well as the fact that the constant can be put in front of the sum: $$A=\sum_{i=1}^n(i^2+3i+4)=\sum_{i=1}^ni^2+3\sum_{i=1}^ni+\sum_{i=1}^n4.$$ Now, using equalities from Theorem 3 we get $$A=\frac{n(n+1)(2n+ 1)}{6}+3\frac{n(n+1)}{2}+4n=\\ n\left(\frac{(n+1)(2n+1)}{6}+\frac{3(n+1)}{2}+4\right)=n\left((n+1)\left(\frac{2n+1}{6}+\frac{3}{2}\right)+4\right)=n\left((n+1)\left(\frac{2n+1}{6}+\frac{3}{2}\right)+4\right)=n\left((n+1)\frac{2n+1+9}{6}+4\right)=n\left(\frac{(n+1)(n+5)}{3}+4\right)=\frac{1}{3}n\left(n^2+6n+17\right).$$
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.