Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 12 - Sequences; Induction; the Binomial Theorem - 12.1 Sequences - 12.1 Assess Your Understanding - Page 808: 75



Work Step by Step

I know that $\sum_{k=1}^{n} (k^2)=\frac{n(n+1)(2n+1)}{6}.$ Hence $\sum_{k=1}^{16} (k^2+4)=\sum_{k=1}^{16} (k^2)+\sum_{k=1}^{16} (4)=\frac{16(16+1)(32+1)}{6}+16(4)=1496+63=1560$
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.