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 - Chapter Review - Review Exercises - Page 839: 14



Work Step by Step

I know that $\sum_{k=1}^{n} (k)=\frac{n(n+1)}{2}.$ Hence $\sum_{k=1}^{40} (-2k+8)=\sum_{k=1}^{40} (-2k)+\sum_{k=1}^{40} (8)=-2\sum_{k=1}^{40} (k)+40(8)=-2\frac{40(40+1)}{2}+320=-1640+320=-1320$
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