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.

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
