Simplification of Sum 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
$$
\forall n \in N : \sum_{i=1}^n (i + 1)i = \frac{1}{3}n(n+1)(n+2). 
$$

Factor out $(n+1)(n+2)$. What's left in each term? What's the sum of those two expressions?
