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).

There are several things you can do. 
$1)$First of all, you might try the induction way. But as you know, in order to do that we first need to have the right side of the formula and we can guess the right side, which is the hard part and sometimes it is too hard to do. Let's do that!
Let $S(n)=\sum_{i=1}^n i(i+1)(i+2)(i+3)$. 
$S(1)=24$, $S(2)=144$, $S(3)=504$. Now, it looks impossible to guess. Therefore, we will try something simpler.
$2)$ This time we will use the formulae we already know such as 
$\sum_{i=1}^n i=\frac{n(n+1)}{2}$, $\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}$, $\sum_{i=1}^n i^3=\frac{n(n+1)}{2}^2$ , and $\sum_{i=1}^n i^4=\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}$. $$S(n)=\sum_{i=1}^n i(i+1)(i+2)(i+3)=\sum_{i=1}^n (i^4+6i^3+11i^2+6i)=\sum_{i=1}^n i^4+   \quad 6\sum_{i=1}^n i^3+11\sum_{i=1}^n i^2+6\sum_{i=1}^n i$$ By using the formulae above we can conclude that $\sum_{i=1}^n i(i+1)(i+2)(i+3)=\frac{1}{5}n(n+1)(n+2)(n+3)(n+4)$ In fact, now i can see that first method can be used as well.( The formula does not look so complex, one can guess, apparently not me!).
As a result, when we have a polynomial we can use the formulae as we have done for this particular example. You can find the formulae here http://math2.org/math/expansion/power.htm. These formulae are easy to prove(by induction), also you might want to derive them as well, which is a little bit harder but something you can try.
I hope this answers your question, TT.
