Algebra and Trigonometry 10th Edition

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

Chapter 11 - Review Exercises - Page 842: 102


True. $\displaystyle \sum_{i=1}^{5} (i^3+2i)=\displaystyle \sum_{i=1}^{5}i^3+\displaystyle \sum_{i=1}^{5}2i$

Work Step by Step

$\displaystyle \sum_{i=1}^{5} (i^3+2i)=1^3+2(1)+2^3+2(2)+3^3+2(3)+4^3+2(4)+5^3+2(5)=(1^3+2^3+3^3+4^3+5^3)+[2(1)+2(2)+2(3)+2(4)+2(5)]=\displaystyle \sum_{i=1}^{5}i^3+\displaystyle \sum_{i=1}^{5}2i$
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.