Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 11 - Chapter Test - Page 843: 6



Work Step by Step

The sum of the squares of the first $n$ integers can be obtained by the following formula: $\frac{n(n+1)(2n+1)}{6}.$ Thus $\sum_{i=1}^{50} (2i^2+5)=\sum_{i=1}^{50} (5)+\sum_{i=1}^{50} (2i^2)=5\cdot50+2\sum_{i=1}^{50} (i^2)=250+2\cdot\frac{50(50+1)(2(50)+1)}{6}=250+2\cdot\frac{50(50+1)(2(50)+1)}{6}=250+85850=86100$
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.