# Simplification of Sum

I can't see how to simplify from step 1 to step 2 in the following example:

1. $$\frac{1}{3}n(n+1)(n+2)+(n+2)(n+1)$$

2. $$(\frac{1}{3}n+1)(n+1)(n+2)$$

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

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

-
Start by rewriting the $(n+2)(n+1)$ term to $1\cdot(n+1)(n+2)$. –  Henning Makholm Oct 9 '12 at 22:24
Yes, I can see now what joriki meant with factor out both bracketed factors from the whole sum. –  Achmed Durangi Oct 9 '12 at 22:38
Step 1.3 makes no sense to me -- is there a typo there? –  joriki Oct 9 '12 at 23:03
My fault. Makes sense now? –  Achmed Durangi Oct 9 '12 at 23:24
@Stephan: It does. –  joriki Oct 9 '12 at 23:58

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