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 299: 58

Answer

$\therefore \sum_{k=1}^{50}\left(-\frac{1}{1+k}+\frac{1}{k}\right)=\frac{50}{51}$

Work Step by Step

We find: $\begin{aligned} \sum_{k=1}^{50}\left(-\frac{1}{1+k}+\frac{1}{k}\right) &=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{50}-\frac{1}{51}\right) \\ &=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{50}-\frac{1}{51}\right) \\ &=-\frac{1}{51}+1 \\ &=\left[\frac{50}{51}\right] \end{aligned}$ $\therefore \sum_{k=1}^{50}\left(-\frac{1}{1+k}+\frac{1}{k}\right)=\frac{50}{51}$
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