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.
 A: $$\sum\limits_{i=1}^{n}{i(i+1)=}\sum\limits_{i=1}^{n}{{{i}^{2}}+\sum\limits_{i=1}^{n}{i=}}\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{3}$$
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
 $$1(2)+2(3)+3(4)+...+k(k+1)+\color{red}{(k+1)(k+2)}=\frac 13 k(k+1)(k+2)+\color{red}{(k+1)(k+2)}$$
$$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\frac13 (k+1)(k+2)(k+3)$$
A: Here is an easy way of going about the inductive step:
\begin{align}
\sum_{i=1}^{k+1}i(i+1)&= \sum_{i=1}^ki(i+1)+(k+1)(k+2)\\[1em]
&= \frac{k(k+1)(k+2)}{3}+(k+1)(k+2)\\[1em]
&= \frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}\\[1em]
&= \frac{(k+1)(k+2)(k+3)}{3}.
\end{align}
See if you can determine how each line was obtained (where the inductive hypothesis was applied, etc.).
A: When $n=1$, it is trival. 
Suppose it is true for $n=k$, ie. $\sum \limits_i^k i(i+1)=\frac{k(k+1)(k+2)}{3}$. 
Consider $n=k+1$. 
$$\begin{align}
\sum \limits_i^{k+1} i(i+1)=&(k+1)(k+2)+\sum \limits_i^{k} i(i+1) \\
=&(k+1)(k+2)+\frac{k(k+1)(k+2)}{3} \\
=&\frac{(k^2+4k+3)(k+2)}{3} \\
=&\frac{(k+1)(k+1+1)(k+1+2)}{3}.
\end{align}
$$
Done!
A: Your proposition is not true.
try $n = 2$
$1\cdot 2 + 2\cdot3 \ne \frac 13 3\cdot4$
Which is why you are having a tough time proving this by induction.
Induction is usefull to prove things, but it doesn't always have a whole lot of insight why things are the way they are.
This is probably what you should have:
$\sum_\limits{k=1}^{n} (k)(k+1) = \frac 13 n(n+1)(n+2)$
base case: 
$n = 1\\
1\cdot2 = \frac13 1\cdot2\cdot 3$
Inductive hypothesis:
$\sum_\limits{k=1}^{n} (k)(k+1) = \frac 13 n(n+1)(n+2)$
Show that
$\sum_\limits{k=1}^{n+1} (k)(k+1) = \frac 13 (n+1)(n+2)(n+3)$ 
based on the inductive hypothesis.
$\sum_\limits{k=1}^{n+1} (k)(k+1)\\
\sum_\limits{k=1}^{n} (k)(k+1) + (n+1)(n+2)\\
\frac 13 n(n+1)(n+2) + (n+1)(n+2)\\
(n+1)(n+2) (\frac 13 n + 1)\\
\frac 13(n+1)(n+2)(n + 3)$
QED
A: Let $ 2 = \frac{1 \cdot 2 \cdot 3}{3} $ be the basis. The inductive step consists of simple distribution of multiplication:
$$
\begin{align}&\frac{n(n+1)(n+2)}{3} + (n+1)(n+2) \\&= \tfrac{1}{3}n^3+2n^2+\tfrac{11}{3}n+2 \\&= \frac{(n + 1)\Big((n+1)+1\Big)\Big((n + 1)+2\Big)}{3}\end{align}
$$
