Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 11 - Sequences; Induction; the Binomial Theorem - Chapter Review - Review Exercises - Page 859: 15



Work Step by Step

We know that $\sum_{k=1}^{n} \ k =\dfrac{n(n+1)}{2}$ Plug in $40$ for $n$ in the given summation formula. Therefore, $\sum_{k=1}^{40} (-2k+8) =\sum_{k=1}^{40} (-2k) +\sum_{k=1}^{40} \ 8 \\ =-2 \sum_{k=1}^{40} k +(40) \ (8) \\=(-2) \times \dfrac{40(40+1)}{2}+320\\=-1640+320 \\=-1320$
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.