College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 9 - Section 9.1 - Sequences - 9.1 Assess Your Understanding - Page 648: 78

Answer

$-2376$

Work Step by Step

... we want to apply (1) $\displaystyle \sum_{k=1}^{n}(ca_{k})=c\sum_{k=1}^{n}a_{k}$, but the index does not start at 1. $\displaystyle \sum_{k=8}^{40}(-3k)=$ (terms from 8 to 40) = (40 terms) - (first 7 terms) $=\displaystyle \sum_{k=1}^{40}(-3k)-\sum_{k=1}^{7}(-3k)$ ... apply formula (1) $=(-3)\displaystyle \sum_{k=1}^{40}k-(-3)\sum_{k=1}^{7}k$ ... apply $6.\displaystyle \qquad \sum_{k=1}^{n}k=1+2+3+\cdots+n=\frac{n(n+1)}{2}$ $=-3\displaystyle \left[\frac{40(40+1)}{2}-\frac{7(7+1)}{2}\right]$ $=-3[820-28]$ $=-2376$
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