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).

Let us conjugate freak_warrior's tip with a telescopic sum to find the closed formula.
Let us set $a_n = n(n+1)(n+2)(n+3) $
You want to find
$$\sum_{i = 1}^{k} a_i $$
Rewrite 
$$a_n = \frac{n(n+1)(n+2)(n+3)(n+4)}{5} - \frac{(n-1)n(n+1)(n+2)(n+3)}{5} $$
One can prove this is true by factoring out $n(n+1)(n+2)(n+3)$.
Set $l_n = \frac{n(n+1)(n+2)(n+3)(n+4)}{5}$ and $r_n = \frac{(n-1)n(n+1)(n+2)(n+3)}{5}$. Now we have
$$a_n = l_n - r_n $$
But as you can see, $r_{n+1} = l_n $ thus
$$\sum_{i = 1}^{k} a_i = \sum_{i = 1}^{k} l_i - r_i = \sum_{i = 1}^{k} r_{i+1} - r_i$$
If you apply the telescopic sum, you get $r_{k+1} - r_1$
The telescopic sum is noticing that two adjacent terms always cancel:
$$\sum_{i = 1}^{k} r_{i+1} - r_i = (r_2 - r_1) + (r_3 - r_2) + \cdots + (r_k - r_{k-1}) + (r_{k+1} - r_k) = r_{k+1} - r_1 $$
Now all you have to do is substitute the desired value of $k $.
