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?

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    $\begingroup$ $\sum_{k=s}^n \binom{k}s=\binom{n+1}{s+1}$. This is the Hockey Stick identity. $\endgroup$ – Mike Earnest Jan 29 '19 at 1:24
  • $\begingroup$ The question is to find such polynomial $F_s(n,k)$ that $\sum_{k}^{n}\binom{k}{s}=\sum_{k}^{n}F_s(n,k)$. From our example it follows that: $F_2(n,k)=k(n-k)$ and $\sum_{k=1}^{n}\binom{k}{2}=\sum_{k=1}^{n}F_2(n,k)$. Can you provide examples for $s>2$? $\endgroup$ – PKK Jan 29 '19 at 1:33
  • $\begingroup$ I understand now. Still, my comment just shows that $P_s(n)$ is nothing mysterious. $\endgroup$ – Mike Earnest Jan 29 '19 at 1:51
  • $\begingroup$ Why don't you mention your previous question math.stackexchange.com/q/2774300 ? $\endgroup$ – Jean Marie Jan 30 '19 at 18:25
  • $\begingroup$ thank you for reminding, by the way these coefficients could be found as \begin{equation} (2k-1)!T(2n,2k)=\frac{1}{r}\sum_{j=0}^{r}(-1)^j\binom{2r}{j}(r-j)^{2n}, \end{equation} where $r=n-k+1$ and $T(2n,2k)$ is central factorial number $\endgroup$ – PKK Jan 30 '19 at 18:30

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

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  • $\begingroup$ Can you show me a direct example for $s=2$ step by step ? The example of $\sum_{k=1}^n\binom{k}{s}$ please $\endgroup$ – PKK Jan 29 '19 at 1:53
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    $\begingroup$ @PetroKolosov When $s=2$, you get $F_s(n,k)=k(n-k)$, just as you had. When $s>2$, the expression $k\binom{n-k}s$ is polynomial in disguise. For example, $F_3(n,k)=k\binom{n-k}2=k(n-k)(n-k-1)/2$. $\endgroup$ – Mike Earnest Jan 29 '19 at 2:36
  • $\begingroup$ Are you sure that $F_3(n,k)=k(n-k)(n-k-1)/2$? For me it gives as follows: $F_s(n,k)=k\binom{n-k}{s-1}$, by the hockey stick pattern: $\binom{n-k}{s-1}=\sum_{j}^{n-k+1}\binom{j}{s-2}|_{s=3}=1/2 (-2 + k - n) (-1 + k - n)$ and multiplication by $k$ gives $F_3(n,k)=1/2 k (-2 + k - n) (-1 + k - n)$. PS and obviously i wrong with limits of hockey stick pattern :) $\endgroup$ – PKK Jan 29 '19 at 3:09
  • $\begingroup$ I've fixed the error above, but still $F_s(n,k)=k\binom{n-k}{s-1}$, by the hockey stick pattern: $\binom{n-k}{s-1}=\sum_{j}^{n-k-1}\binom{j}{s-2}|_{s=3}=1/2 (k - n) (1 + k - n)$ and multiplication by $k$ gives $F_3(n,k)=1/2 k (k - n) (1 + k - n)$ is different from your example of $F_3(n,k)$ $\endgroup$ – PKK Jan 29 '19 at 3:18
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    $\begingroup$ $k(n-k)(n-k-1)/2=\frac12k(k-n)(1+k-n)$. Your expression and mine for $F_3(n,k)$ are the same. All you did was expand $\binom{n-k}{s-1}$ using the HS identity, and then collapse it using the same identity. @PetroKolosov $\endgroup$ – Mike Earnest Jan 29 '19 at 3:37

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

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

Going back to the favored solution:


If you wanted the thing on the right to be


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

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

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.

If we want the right-hand side to be


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 $s=n+1-s.$

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