Answer
$\frac{98}{303}$
Work Step by Step
$\frac{2}{x(x+2)}=\frac{(x+2)-x}{x(x+2)}=\frac{1}{x}-\frac{1}{x+2}$, thus
$\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}=(\frac{1}{1}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+...+(\frac{1}{99}-\frac{1}{101})=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}$
