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