Answer
See below.
Work Step by Step
1. For $n=1$, we have $LHS=\frac{1}{1\cdot2}=\frac{1}{2}$ and $RHS=\frac{1}{1+1}=\frac{1}{2}=LHS$, thus P(1) is true.
2. Assume P(k) ($k\gt1$) is true, we have:
$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{k\cdot(k+1)}=\frac{k}{k+1}$
3. For $n=k+1$, we have:
$LHS=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{k\cdot(k+1)}+\frac{1}{(k+1)\cdot(k+2)}=\frac{k}{k+1}+\frac{1}{(k+1)\cdot(k+2)}=\frac{k(k+2)+1}{(k+1)\cdot(k+2)}=\frac{(k+1)^2}{(k+1)\cdot(k+2)}=\frac{k+1}{k+2}=RHS$
4. Thus $P(k+1)$ is also true and we have proved the statement by mathematical induction.