Prove by induction:1*2+2*3+3*4+...+n(n+1)=n(n+1)(n+2)/3
Base case: $\displaystyle n=1$
$\displaystyle 1 \cdot 2 = \frac{1 \cdot 2 \cdot 3}{3} = 1 \cdot 2$
Assume correctness for $\displaystyle n=k$. Prove for $\displaystyle n=k+1$:
LHS:
$\displaystyle 1 \cdot 2 + 2 \cdot 3 + ... + n(n+1) + (n+1)(n+2) = \frac{n(n+1)(n+2)}{3} + (n+1)(n+2) =$ $\displaystyle \frac{n(n+1)(n+2)}{3} + \frac{3(n+1)(n+2)}{3} =$ $\displaystyle \frac{n(n+1)(n+2) + 3(n+1)(n+2)}{3} = \frac{(n+1)(n+2)(n+3)}{3}$
RHS:
$\displaystyle \frac{(n+1)(n+2)(n+3)}{3}$