claim: $\sum _{i\:=\:0}^n\:i\left(i+1\right)\:=\:\frac{n\left(n+1\right)\left(n+2\right)}{3}$
base case n = 0
$\frac{0\left(0+1\right)\left(0+2\right)}{2}$ ; $0\:=\:0$
inductive hyp (n = k):
$\sum _{i=0}^k\:i\left(i+1\right)\:=\:\frac{k\left(k+1\right)\left(k+2\right)}{3}$
to show: n = k+1
$\sum _{i=0}^k\:i\left(i+1\right)\:=\:\frac{\left(k+1\right)\left(k+1+1\right)\left(k+1+2\right)}{3}$
$\sum _{i=0}^k\:i\left(i+1\right)\:=\:\frac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{3}$
not sure what to do next
For the case $n=k+1$:
$$\sum_{i=0}^{k+1}\:i\left(i+1\right)\:=\sum_{i=0}^{k}\:i\left(i+1\right)+(k+1)(k+2)$$
Using inductive hyp we get:
$$\sum_{i=0}^{k+1}\:i\left(i+1\right)\:=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)=$$
$$(k+1)(k+2)\left(\frac{k}{3}+1\right)=\frac{(k+1)(k+2)(k+3)}{3}$$