TITLE: Proof for sum of product of four consecutive integers
QUESTION [5 upvotes]: I had to prove that
$(1)(2)(3)(4)+\cdots(n)(n+1)(n+2)(n+3)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}$
This is how I attempted to do the problem:
First I expanded the $n^{th}$ term:$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$.
So the series $(1)(2)(3)(4)+\cdots+(n)(n+1)(n+2)(n+3)$ will be equal to
$(1^4+2^4+\cdots+n^4)+6(1^3+2^3+\cdots+n^3)+11(1^2+2^2+\cdots+n^2)+6(1+\cdots+n)$
$=\frac{n(n+1)(2n+1)(3n^2-3n-1)}{30}+6*\frac{n^2(n+1)^2}{4}+11*\frac{(n+1)(2n+1)(n)}{6}+6*\frac{n(n+1)}{2}$
I tried to factor out $\frac{n(n+1)}{5}$ and tried to manipulate the expressions in ways which further complicated things. How should I proceed? Can anyone please give me some clues? Any help is appreciated.

REPLY [3 votes]: $$\sum_{k=1}^n k(k+1)(k+2)(k+3)$$
$$=\sum_{k=1}^n 4!\binom{k+3}{4}$$
$$=4!\sum_{k=1}^n \binom{k+3}{4}$$
Using the hockey-stick identity:
$$=4!\binom{k+4}{5}$$
$$=\frac{k(k+1)(k+2)(k+3)(k+4)}{5}$$