Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.4 The Definition Of Area As A Limit; Sigma Notation - Exercises Set 4.4 - Page 298: 20

Answer

$-n+4$

Work Step by Step

Remember that: $\sum_{k=1}^{n} 1=n$ and $\sum_{k=1}^{n} k=\frac{(1+n)n}{2}$ We find: $\sum_{k=1}^{n}\left(-\frac{2 k}{n}+\frac{5}{n}\right)=-\frac{2}{n} \sum_{k=1}^{n} k+\frac{5}{n} \sum_{k=1}^{n} 1=\frac{5}{n} \cdot n-\frac{2}{n} \cdot \frac{(1+n)n}{2}=-(1+n)+5=5-n-1=-n+4$
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