College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 5 - Systems of Equations and Inequalities - Exercise Set 5.3 - Page 550: 51



Work Step by Step

$\frac{1}{x(x+1)}=\frac{(x+1)-x}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$, thus $\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+...+(\frac{1}{99}-\frac{1}{100})=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}$
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