# Concerning the identity in sums of Binomial coefficients [closed]

Consider 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.) Suppose there exists the representation 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?

## closed as off-topic by darij grinberg, GH from MO, David Handelman, Brendan McKay, Mark WildonJan 29 at 3:33

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• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – darij grinberg, GH from MO, David Handelman, Brendan McKay, Mark Wildon
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• This is not a research level question. At any rate, it is straightforward to see that $\sum_{k=1}^n\binom{k}{s}=\binom{n+1}{s+1}$. Indeed, choosing $s+1$ numbers out of $\{1,2,\dots,n+1\}$ is the same as choosing the largest number first (call this $k+1$), and then choosing the other $s$ numbers from $\{1,2\dots,k\}$. Please ask non-research level questions at math.stackexchange.com – GH from MO Jan 29 at 0:18
• The question is revised according to your notice. The sense is following: Can we always find the polynomials $F_s(n,k)$, such that $\sum_{k}^{n}\binom{k}{s}=\sum_{k}^{n}F_s(n,k)$ ? – Petro Kolosov Jan 29 at 0:43
• Well, the sum is a single polynomial $\binom{n+1}{s+1}$. It is of degree $s+1$ in $n$. – GH from MO Jan 29 at 0:50
• What does $\sum_k^s$ mean? Anyway, this is moving to math.stackexchange after one more vote, so no need to repost it there yourself :) – darij grinberg Jan 29 at 0:50
• Your question doesn't make sense except that which GH already answered. – Brendan McKay Jan 29 at 0:52