\begin{equation*} \sum_{r=1}^{n} r^\alpha (n-r)^\beta = ? \end{equation*} where $\alpha$ and $\beta$ are natural numbers
Background:
Here's something I wish to share with the Math SE community, a very interesting combinatorial proof, that I came across sometime ago. I strongly believe,that this approach could lead to a possible generalisation of similar summations.
Those who wish to read the following warm-up proofs (I thought it's a good idea to share what motivated me towards asking the problem I intend to ask), may do so while enjoying the little snack for the mind - the art of combinatorial proof. Others, may skip directly to the question I intend to ask, right after these proofs.
\begin{equation*} 1 + 2 + 3 + \cdots + n = {n+1 \choose 2} \end{equation*}
The above identity is a very familiar one, apparently one that Gauss came up with when his teacher asked him to sum up numbers from 1 to 100 (supposed to be punishment). Of course, his proof is well known - writing the sequence in the reverse order and summing it with the original one greatly reduces computational effort.
Now, what's interesting is the combinatorial approach, which I summarise as follows (for people who may not be familiar with it)-
In general, to give a combinatorial proof for a binomial identity, say A=B you do the following:
Find a counting problem you will be able to answer in two ways. Explain why one answer to the counting problem is A. Explain why the other answer to the counting problem is B. Since both A and B are the answers to the same question, we must have A=B.
The tricky thing is coming up with the question. This is not always obvious, but it gets easier the more counting problems you solve. You will start to recognize types of answers as the answers to types of questions. More often what will happen is you will be solving a counting problem and happen to think up two different ways of finding the answer. Now you have a binomial identity and the proof is right there. The proof is the problem you just solved together with your two solutions.
Now, coming to the proof of the above-stated identity-
Consider the question:"How many subsets of $A = {1,2,3, \ldots, n+1}$ contain exactly two elements?" Answer 1: We must choose 2 elements from $n+1$ in set A, so there are exactly ${n+1 \choose 2}$ subsets.
Answer 2: We break this question down into cases, based on what the larger of the two elements in the subset is. The larger element can't be 1, since we need at least one element smaller than it.
Larger element is 2: there is 1 choice for the smaller element.
Larger element is 3: there are 2 choices for the smaller element.
Larger element is 4: there are 3 choices for the smaller element. And so on. When the larger element is $n+1$, there are $n$ choices for the smaller element. Since each two element subset must be in exactly one of these cases, the total number of two element subsets is $1 + 2 + 3 + \cdots + n\text{.}$
Answer 1 and answer 2 are both correct answers to the same question, so they must be equal. Therefore, \begin{equation*} 1 + 2 + 3 + \cdots + n = {n+1 \choose 2} \end{equation*}
To drive y'all another step closer to my question, here's another (just one) combinatorial proof: \begin{equation*} 1(n) + 2(n-1) + 3 (n-2) + \cdots + (n-1) 2 + n(1) = {n+2 \choose 3}. \end{equation*}
Consider the question “How many 3-element subsets are there of the set $\{1,2,3,\ldots, n+2\}\text{?}$"
Answer 1: We must select 3 elements from the collection of $n+2$ elements. This can be done in ${n+2 \choose 3}$ ways.
Answer 2: Break this problem up into cases by what the middle number in the subset is. Say each subset is $\{a,b,c\}$ written in increasing order. We count the number of subsets for each distinct value of $b$. The smallest possible value of $b$ is $2$ and the largest is $n+1$
When $b=2$, there are $1(n)$ subsets: 1 choice for $a$ and $n$ choices (3 through $n+2$) for $c$
When $b=3$, there are $2(n-1)$ subsets: 2 choices for $a$ and $n-1$ choices for $c$
When $b=4$, there are $3(n-2)$ subsets: 3 choices for $a$ and $n-2$ choices (3 through $n+2$) for $c$
And so on. When $b=n+1$ there are $n$ choices for $a$ and only 1 choice for $c$, so, $n(1)$ subsets.
Therefore, the total number of subsets is $1(n) + 2(n-1) + 3 (n-2) + \cdots + (n-1) 2 + n(1)$.
Since Answer 1 and Answer 2 are answers to the same question, they must be equal.
Therefore, \begin{equation*} 1(n) + 2(n-1) + 3 (n-2) + \cdots + (n-1) 2 + n(1) = {n+2 \choose 3}. \end{equation*}
This, sort of hints at a pattern, in summations involving r and n-r multiplied, raised to different powers(r varying from 1 to n). Finally, my question is: \begin{equation*} \sum_{r=1}^{n} r^\alpha (n-r)^\beta = ? \end{equation*} where $\alpha$ and $\beta$ are natural numbers
I need help not only in solving for the answer to the above summation in terms of $\alpha$ and $\beta$, but also figuring out a combinatorial approach. The power of the art of combinatorial proof convinces me that the above summation, a generalisation of what we have seen till now - can be possibly evaluated using combinatorial techniques/arguments.
Please help, thanks a lot.
Edit: Something just struck me. Can we look at possible recursion, for the given sum? We know how to solve the easy cases, like ($\alpha$,$\beta$)= (0,1) or (1,0). Is it possible to find a recursion of S($\alpha$,$\beta$) in terms of S($\alpha$-1,$\beta$), or S($\alpha$,$\beta$-2), etc, where S denotes the required sum with parameters ($\alpha$,$\beta$)