How can I derive what is $1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$ ?? I'd like to evaluate the series $$1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$$
Since I am a high school student, I only know how to prove such formula's (By principal of mathematical induction). I don't know how to find result of such series.
Please help. I shall be thankful if you guys can provide me general solution (Since I have been told that there exist a general solution by my friend who gave me this question).

In general, we have
$$\sum_{k=1}^nk(k+1)(k+2)\dots(k+p)=\frac{n(n+1)(n+2)\dots(n+p+1)}{p+2}$$
Prove by induction,
$$\sum_{k=1}^{n+1}k(k+1)(k+2)\dots(k+p)\\=(n+1)(n+2)\dots(n+p+1)+\sum_{k=1}^nk(k+1)(k+2)\dots(k+p)\\=(n+1)(n+2)\dots(n+p+1)+\frac{n(n+1)(n+2)\dots(n+p+1)}{p+2}\\=\frac{\color{#034da3}{(p+2)}\color{#ee8844}{(n+1)(n+2)\dots(n+p+1)}+\color{#034da3}n\color{#ee8844}{(n+1)(n+2)\dots(n+p+1)}}{p+2}\\=\frac{\color{#ee8844}{(n+1)(n+2)\dots(n+p+1)}\color{#034da3}{(n+p+2)}}{p+2}$$
And easy enough to check for $n=1$ to see it is true.
$$1\times2\times3\times\ldots\times(1+p)=\frac{1\times2\times3\times\ldots\times(1+p)\times\require{cancel}\cancel{(2+p)}}{\cancel{p+2}}$$

This is slightly off, since my sum ends at $n(n+1)(n+2)(n+3)$, while you end off at $(n-3)(n-2)(n-1)n$.  To readjust, have $n=p-3$ in my sum and it will become yours.
