<s>[INST] Proof for sum of product of four consecutive integers 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. 
 [/INST] If I were you, I would try induction. First establish that it is true for $n=1$:
$$1*2*3*4=24=\frac{1(2)(3)(4)(5)}{5}$$
Then use induction by first assuming that, for some $k$,
$$1*2*3*4+...+k(k+1)(k+2)(k+3)=\frac{k(k+1)(k+2)(k+3)(k+4)}{5}$$
Then add $(k+1)(k+2)(k+3)(k+4)$ to both sides:
$$1*2*3*4+...+(k+1)(k+2)(k+3)(k+4)=\frac{k(k+1)(k+2)(k+3)(k+4)}{5}+(k+1)(k+2)(k+3)(k+4)$$
If you want to try this on your own, stop reading here and try the induction yourself.
Factor out $(k+1)(k+2)(k+3)$ from the right side:
$$1*2*3*4+...+(k+1)(k+2)(k+3)(k+4)=(k+1)(k+2)(k+3)(\frac{k(k+4)}{5}+(k+4))$$
$$1*2*3*4+...+(k+1)(k+2)(k+3)(k+4)=(k+1)(k+2)(k+3)(\frac{(k+5)(k+4)}{5})$$
$$1*2*3*4+...+(k+1)(k+2)(k+3)(k+4)=\frac{(k+1)(k+2)(k+3)(k+4)(k+5)}{5}$$
And there you have it.
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