Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.3 - Partial Fractions - Exercise Set - Page 842: 52


The simplified solution is $\frac{1}{x}-\frac{1}{x+2}$ and the sum is $\frac{100}{101}$.

Work Step by Step

Let us consider the rational expression. $\begin{align} & \frac{2}{x\left( x+2 \right)}=\frac{A}{x}+\frac{B}{x+2} \\ & =\frac{A\left( x+2 \right)+Bx}{x\left( x+2 \right)} \end{align}$ $2=A\left( x+2 \right)+Bx$ (I) Substitute the value of $x=-2$ in the equation (I), $\begin{align} & 2=B\left( -2 \right) \\ & B=-1 \end{align}$ Again, putting the value of $x=0$ in (I), $\begin{align} & 2=A\left( 2 \right) \\ & A=1 \end{align}$ Therefore, $\frac{2}{x\left( x+2 \right)}=\frac{1}{x}-\frac{1}{x+2}$ is the simplified form of the result. And use the above result to find the sum of the series, $\begin{align} & \frac{2}{1\cdot 3}+\frac{2}{3\cdot 5}+\frac{2}{5\cdot 7}+\cdots +\frac{2}{99\cdot 101}=\left( \frac{1}{1}-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{5} \right)+\left( \frac{1}{5}-\frac{1}{7} \right)+\cdots +\left( \frac{1}{99}-\frac{1}{101} \right) \\ & =\frac{1}{1}-\frac{1}{101} \\ & =\frac{100}{101} \end{align}$ Thus, the sum is $\frac{100}{101}$.
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