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.
Can you prove that $(n-3)(n-2)(n-1)(n) = \frac{(n-3)(n-2)(n-1)(n)(n+1)}{5} - \frac{(n-4)(n-3)(n-2)(n-1)(n)}{5}?$
From there you can substitute different values of $n$ and derive a formula for your expression, and the final summation will be given by
$$ \frac{(n-3)(n-2)(n-1)(n)(n+1)}{5}$$
Following up from Yves Daoust's answer, assume that our sum is a polynomial. Clearly it must be a polynomial of degree $5$ with $1/5$ coefficient because
$$P(n)=\sum_{k=1}^nk(k+1)(k+2)(k+3)\sim\int_1^nx^4+\mathcal O(x^3)dx=\frac15x^5+\mathcal O(x^4)$$
Notice that if we have
$$P(1)=24\qquad P(n)=n(n+1)(n+2)(n+3)+P(n-1)$$
Then
$$P(-4)=P(-3)=P(-2)=P(-1)=P(0)=0$$
Meaning, by the fundamental theorem of algebra,
$$P(n)=\frac{n(n+1)(n+2)(n+3)(n+4)}5$$
The product can also be identified as parts of binomials, as $$ \binom{n}4=\frac{n(n-1)(n-2)(n-3)}{4!} $$ From the Pascal triangle identity we know that $\binom{n+1}{k+1}=\binom{n}{k+1}+\binom{n}{k}$ where we can put the same denominors on one side $$ \binom{n}{k}=\binom{n+1}{k+1}-\binom{n}{k+1} $$
so that finally $$ \sum_{n=k}^N\binom{n}{k}=\sum_{n=k}^N\binom{n+1}{k+1}-\binom{n}{k+1} $$ is found to be a telescoping sum.
$$\color{blue}{\sum_{r=4}^{n} (r-3)(r-2)(r-1)r}=4!\sum_{r=4}^{n}\binom r4=4!\binom {n+1}5=\color{red}{\frac{(n-3)(n-2)(n-1)n(n+1)}5}$$
Using the Pochhammer notation for the rising factorial, the above may be written as
$$\color{blue}{\sum_{r=4}^n (r-3)^{\overline{4}}}=\color{red}{\frac{(n-3)^\overline{5}}5}$$
It is interesting to note the similarity between this and the standard integral of a power
$$\int_0^n r^4 \;dr=\frac{n^5}5$$
General Case
The above can be generalized to
$$\sum_{r=m+1}^n (r-m)^{\overline{m+1}}=\frac{\;\;(n-m)^{\overline{m+2}}}{m+2\;\;\;}$$
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 $.
The answer must be a quintic polynomial $P(n)$, because $P(n)-P(n-1)$ is a quartic polynomial.
Then it suffices to choose $5$ values of $n$, evaluate the corresponding sums and write the Lagrangian interpolation polynomial, which is guaranteed to be unique.
You can shorten the computation a little by noting that the coefficient of $n^5$ must be $1/5$ (as $n^5-(n-1)^5=5n^4+\cdots$), and by the shifting $m=n-3$ use the fact that $P(3)=0$ to get rid of the independent term.
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
$$\sum_{k=4}^{n+4}k^{\underline 4}=\frac15(n+5)^{\underline 5}$$
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.
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
