Let $$f(n,k) = \sum\limits_{j = - k}^k {{{( - 1)}^{k - j}}} \binom{n-j}{k-j}\binom{n+j}{k+j}.$$ Then $f(n,k)=\binom{n}{k}$ because it satisfies $f(n,k)=f(n-1,k)+f(n-1,k-1)$ and the obvious boundary values.

Let $ {n\brack {k}}$ be a $q-$binomial coefficient. I want to know if there is a similar proof for the identity $$f(n,k,q) = \sum\limits_{j =-k}^k {(-1)^{k-j}} q^{\binom{j}{2}+\binom{k+1}{2}-{(n+1)k}} {{n-j}\brack{k-j}}{{n+j}\brack{k+j}}={n\brack{k}}.$$ There is an easy computer proof using the $q-$Zeilberger algorithm and perhaps it also follows from a simple $q-$ hypergeometric summation. But I am interested in a direct proof.

  • 1
    $\begingroup$ If the $q$-identity is true then the recurrence $f(n,k,q) = q^kf(n-1,k,q)+f(n-1,k-1,q)$ must be satisfied. It might help if you were to display the steps of the proof of the original (non-$q$) identity and point out which step doesn't $q$-ify easily. $\endgroup$ Feb 23, 2015 at 21:56
  • $\begingroup$ @Timothy:My proof for $q=1$ uses the fact that $f(n,k)=g(n,k)= \sum\limits_{j = - k}^k {{{( - 1)}^{k - j}}}\binom{n-j-1}{k-j}\binom{n+j}{k+j}.$ Then it is easy to verify that $g(n+1,k)=f(n,k)+f(n,k-1).$ In the general case you get instead $f(n,k,q)=g(n,k,q)= \sum\limits_{j =-k}^k {(-1)^{k-j}} q^{\binom{j+1}{2}+\binom{k+1}{2}-{(n+1)k}}{{n-1-j}\brack{k-j}}{{n+j}\brack{k+j}}$ and in $g(n+1,k,q)-f(n,k,q)$ you get the terms ${{n+1+j}\brack{k+j}}-q^k{{n+j}\brack{k+j}}$ which are no longer multiples of ${n+j}\brack{k+j-1}$ as for $q=1.$ $\endgroup$ Feb 24, 2015 at 10:06

1 Answer 1


At first, we use a formula $\binom{u}{m}=(-1)^m\binom{m-u-1}{m}$. Thus $$\binom{n\pm j}{k\pm j}=(-1)^{k-j}\binom{k-n-1}{k\pm j}.$$ Denote $k-n-1=x$, and $k-j$ by $s\in \{0,\dots,2k\}$, we have to prove that $$ \sum_{s=0}^{2k} (-1)^s \binom{x}{s}\binom{x}{2k-s}=(-1)^k \binom{x}{k}. $$ LHS counts the coefficient of $t^{2k}$ in $(1-t)^x\cdot (1+t)^x$, as is seen immediately immediately from expanding both multiples $(1\pm t)^x=\sum (\pm 1)^s\binom{x}{s} t^s$.

Now we have to $q$-ify this argument, and this seems to be dooable. Indeed, instead of $(1\pm t)^x$ we consider the $q$-power $(1\pm t)(1\pm qt)\dots (1\pm q^{x-1}t)$. The coefficients are known by $q$-binomial theorem, and their values allow to define this $q$-power for non-natural $x$ (each specific coefficient is a polynomial in $q^x$). The product of two such $q$-powers is $q^2$-power of $t^2$, which again expands by $(q^2)$-binomial theorem.

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    $\begingroup$ Thank you for the idea of negating the upper index. Doing this in the general case gives the identity $\sum\limits_{j =-k}^k {(-1)^{k-j}} q^{\frac{j(3j-1)}{2}} {{k-n-1}\brack{k-j}}{{k-n-1}\brack{k+j}}= q^{\binom{k+1}{2}-{(n+1)k}}{{n}\brack{k}}={{k-n-1}\brack{k}},$ which has a simple elementary proof (cf. e.g. arxiv.org/abs/math/0701802). $\endgroup$ Feb 24, 2015 at 12:33

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