Proving that $1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 } $ by induction 
Prove that 
  $$1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right)  }{ 3 }. $$

I can get to $1/3(k+1)(k+2) + (k+1)(k+2)$ but then finishing off  and rearranging the problem to $1/3(k+2)(k+3)$ is where I always get stuck.

$$\sum\limits_{i=1}^{n}{i(i+1)=}\sum\limits_{i=1}^{n}{{{i}^{2}}+\sum\limits_{i=1}^{n}{i=}}\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{3}$$
for $n=1$we have $1(2)=\frac{1(1+1)(1+2)}{3}$
let 
$$1(2)+2(3)+3(4)+...+k(k+1)=\frac 13 k(k+1)(k+2)$$
we show
$$1(2)+2(3)+3(4)+...+(k+1)(k+2)=\frac 13 (k+1)(k+2)(k+3)$$
Proof
 $$1(2)+2(3)+3(4)+...+k(k+1)+\color{red}{(k+1)(k+2)}=\frac 13 k(k+1)(k+2)+\color{red}{(k+1)(k+2)}$$
$$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\frac13 (k+1)(k+2)(k+3)$$
