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 297: 2

Answer

(a) $\frac{(1+n)n}{2}$ (b) $n+(1+n)3 n$ (c) $\frac{n(1+n)(1+2 n)}{6}$

Work Step by Step

Using theorem 5.4 .2: (a) $\sum_{k=1}^{n} k=\frac{(1+n)n}{2}$ Using theorems 5.4 .1 and 5.4 .2: (b) $\sum_{k=1}^{n}(1+6 k)=6 \sum_{k=1}^{n} k+\sum_{k=1}^{n} 1$ $=6 \frac{(1+n)n}{2}+n=n+(1+n)3 n$ $(\mathrm{c}) \sum_{k=1}^{n} k^{2}=\frac{(1+n)(1+2 n)n}{6}$
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