Concerning the identity in sums of Binomial coefficients

Question

Let be the following identity $$\sum_{k=1}^{n}\binom{k}{2}=\sum_{k=0}^{n-1}\binom{k+1}{2}=\sum_{k=1}^{n}k(n-k)=\sum_{k=0}^{n-1}k(n-k)=\frac16(n+1)(n-1)n$$ As we can see the partial sums of binomial coefficients are expressed in terms of $3$-rd order polynomial $P_3(n)$, where $n$ is variable of upper bound of summation. We assume that order of resulting polynomial $P_3(n)$ depends on subscript of binomial coefficient being summed up (in our case the order of polynomial is $3=2+1$, where $2$ is subscript of bin. coef.)

The question: Does there exist a generalized method to represent the sum of binomial coefficients $\sum_{k}^{n}\binom{k}{s}$ in terms of certain polynomials $P_{s+1}(n)=\sum_{k}^{n} F_s(n,k)$ for every non-negative integer $s$? I.e can we always find the function $F_s(n,k)$, such that $\sum_{k}^{n}\binom{k}{s}=\sum_{k}^{n}F_s(n,k)$ ? We assume that order of polynomial is $s+1$ by means of example above.

The sub-question: (In case of positive answer to the first question.) If there exists a method to represent the sums of bin. coef. in terms of polynomials in $n$, how do summation limits of the $\sum_{k}^{n}\binom{k}{s}$ implies to the form of polynomial $P_{s+1} (n)$ exactly?

Answer 1

$$ \sum_{k=0}^n\binom{k}s=\sum_{k=0}^n\sum_{i=0}^{k-1}\binom{i}{s-1}=\sum_{i=0}^{n-1}\sum_{k=i+1}^n\binom{i}{s-1}=\sum_{i=0}^{n-1}(n-i)\binom{i}{s-1}=\sum_{i=1}^ni\binom{n-i}{s-1} $$ In other words, you can let $F_s(n,k)=k\binom{n-k}{s-1}$, and you will have $\sum_{k=0}^n \binom{k}s=\sum_{k=0}^{n}F_s(n,k)$.

Answer 2

I would say that you have a good answer already. But their are other possible answers which seem reasonable. Further restriction might force the favored solution above.

In the case $k=3$ (which is the only one I will discuss in any detail)

$$\sum_{s=1}^n\binom{s}3= \\ \sum_{s=1}^ns\binom{n-s}2=\sum_{s=1}^n(n-s)\binom{s}2=\frac14\sum_{s=1}^n{s(n-s)(n-2)}=\frac1{24}\sum_{s=1}^n{(n+1)(n-1)(n-2)}$$

It is easy to see how to generalize these to other $k.$

The first three belong to the infinite family

$$\frac14\sum_{s=1}^n{s(n-s)(\alpha n-(2\alpha -2)s-2)} \tag{*}$$

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Going back to the favored solution:

$$\sum_{s=1}^n\binom{s}3=\sum_{s=1}^n\frac{s^3}{6}-\frac{s^2}{2}+\frac{s}{3}=\sum_{s=1}^n\frac{n^2s}2-n{s}^{2}+\frac{s^3}2-\frac{ns}{2}+\frac{s^2}2.$$

If you wanted the thing on the right to be

$$\sum_{s=1}^nA{n^2s}+Bn{s}^{2}+C{s^3}+D{ns}+E{s^2}$$

Then you do need to have $E=\frac12$ But the rest have two degrees of freedom

$$C=-2A-\frac43B \ \ \ \ \ \ \ D=A-\frac13B-\frac23$$

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For further restriction we might want to have $D=-E=-\frac12$ so than $A{n^2s}+Bn{s}^{2}+C{s^3}+D{ns}+E{s^2}=0$ when $s=n$

In this case the summand factors (of course) giving the family $(*)$ above.

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If we want the right-hand side to be

$$K\sum_{s=1}^ns(An+(1-A)s+B)(Cn+(1-C)s+D)$$

It is possible to work out the requirements. I came up with $6$ families of solutions. Of course the set of solutions is invariant under switching the two terms and/or substituting $s=n+1-s.$