## Precalculus (6th Edition) Blitzer

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