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).

As shown in this post,
$$ \sum_{k=1}^n x^k = x \sum_{k=1}^{n}  \binom{n}{k} (x-1)^{k-1}$$
Differentiate five times:
$$ \sum_{k=1}^n k (k-1)(k-2)(k-3) x^{k-4} = \frac{d^5}{dx^5}  \left[ x \sum_{k=1}^{n}  \binom{n}{k} (x-1)^{k-1} \right] \tag{1}$$
For left side,
$$ \sum_{k=1}^n k (k-1)(k-2)(k-3)(k-4) x^{k-4}$$
We can start this sum:
$$ \sum_{k=5}^n (k-1)(k-2)(k-3) x^{k-4}$$
And, I can sub $k-5 =u$, this turns into:
$$ \sum_{u=0}^{n-5} (u+4)(u+3)(u+2)(u+1)x^u$$
And putting this back in(1):
$$ \sum_{u=0}^{n-5} (u+4)(u+3)(u+2)(u+1)x^u= \frac{d^5}{dx^5}  \left[ x \sum_{k=1}^{n}  \binom{n}{k} (x-1)^{k-1} \right] \tag{1}$$
Now shift $ n \to n +5$, this leads to:
$$ \sum_{u=0}^{n} (u+4)(u+3)(u+2)(u+1)x^u= \frac{d^5}{dx^5}  \left[ x \sum_{k=1}^{n+5}  \binom{n+5}{k} (x-1)^{k-1} \right] \tag{1}$$
Evaluate RHS using leibniz product rule and take limit on both side as $x \to 1$, you get the formula
