I can't see how to simplify from step 1 to step 2 in the following example:
$$ \frac{1}{3}n(n+1)(n+2)+(n+2)(n+1) $$
$$ (\frac{1}{3}n+1)(n+1)(n+2) $$
Thanks to the answers this is how I got from 1 to 2:
1.1 $$ \frac{1}{3}n(n+1)(n+2)+1(n+2)(n+1) $$ 1.2 $$ (n+2)\left(\frac{1}{3}n(n+1)+1(n+1)\right) $$ 1.3 $$ \left((n+1)(\frac{1}{3}n+1)\right)(n+2) $$ Then you get to step 2. Or factor out both (n+1) and (n+2) from the whole sum at once: $$ (n+1)(n+2)\left((\frac{1}{3}n+1)\right) $$
In case you wonder why all this - now I can show that
$$ \sum_{i=1}^{n+1} (i + 1)i = \left(\sum_{i=1}^n (i + 1)i\right) + (n+2)(n+1) $$
$$ = \frac{1}{3}(n+1)(n+2)(n+3) $$
which should proof (by using mathematical induction) that
$$ \forall n \in N : \sum_{i=1}^n (i + 1)i = \frac{1}{3}n(n+1)(n+2). $$