Answer
See below.
Work Step by Step
1. Test for $n=1$, we have $LHS=1\cdot2=2$ and $RHS=\frac{1}{3}(1)(1+1)(1+2)=2$, thus it is true.
2. Assume the formula is true for $n=k$, we have
$1\cdot2+2\cdot3+3\cdot4+...+k(k+1)=\frac{1}{3}(k)(k+1)(k+2)$
3. For $n=k+1$, we have $LHS=1\cdot2+2\cdot3+3\cdot4+...+k(k+1)+(k+1)(k+2)=\frac{1}{3}(k)(k+1)(k+2)+(k+1)(k+2)=\frac{1}{3}(k+1)(k+2)(k+3)=RHS$
4. We conclude that the formula is true for any $n$.