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).

Using finite calculus we have that
$$\sum k^{\underline 4}\delta k=\frac{k^{\underline 5}}{5}+C$$
where $k^{\underline 4}=k(k-1)(k-2)(k-3)$ is a falling factorial. Then taking limits
$$\sum_{k=m}^nk^{\underline 4}=\sum\nolimits_m^{n+1}k^{\underline 4}\delta k=\frac{k^{\underline 5}}{5}\bigg|_m^{n+1}$$
The standard case is
$$\sum_{k=4}^{n+4}k^{\underline 4}=\frac15(n+5)^{\underline 5}$$
