Let ${\left( {a;q} \right)_n}=\prod\limits_{j = 0}^{n - 1} {(1-{q^j}a} )$ and let $ {{n}\brack{k}}_q$ denote a $q-$binomial coefficient.

I am interested in $q-$analogues of the identity $ \sum\limits_{j = - k}^k {{{( - 1)}^{ j}}}\binom{n}{k-j}\binom{n}{k+j}=\binom{n}{k}$ where the left-hand side has the form $\sum\limits_{j =-k}^k {(-1)^{j}} q^{a(j)} {{n}\brack{k-j}}_q{{n}\brack{k+j}}_q $ and the right-hand side gives a simple result.

It is known that

for $a(j)= {\frac{j(3j-1)}{2}}$ the right-hand side is ${{n}\brack{k}}_q ,$

for $a(j)=j^2$ the right-hand side is ${{n}\brack{k}}_{q^2}$

and for $a(j)=\binom{j}{2}$ it is $q^{nk-k^2}{{n}\brack{k}}_{q}. $

Computer experiments suggest also the following evaluations: The right-hand side is $(q^{k}+q^{n-k}-q^n){{n}\brack{k}}_{q}$ for $a(j)=3\binom{j}{2}$

and $q^k {{n}\brack{k}}_{q}{\frac{(-q^{n-k+1};q)_{k-1}(1+q^{n-2k}) }{(-q;q)_k}}$ for $a(j)=2\binom{j}{2}.$

I would be interested if the last two results are special cases of some $q-$hypergeometric summation or if there is a direct proof known.

  • $\begingroup$ Are there similar results also for $\binom{j+1}{2}$ instead of $\binom{j}{2}$ ? $\endgroup$
    – F. C.
    Apr 24, 2015 at 20:15
  • $\begingroup$ @F.C. By symmetry the above mentioned sums remain unchanged by changing $a(j)$ to $a(-j)$. Since $\binom{-j}{2}=\binom{j+1}{2}$ the results are the same for $\binom{j+1}{2}.$ But there are changes if you consider the related sums $r(n,k,a(j))=\sum\limits_{j =-k}^k {(-1)^{j}} q^{a(j)} {{n}\brack{k-j}}_q{{n+1}\brack{k+j}}_q $ and $s(n,k,a(j))=\sum\limits_{j =-k}^k {(-1)^{j}} q^{a(-j)} {{n}\brack{k-j}}_q{{n+1}\brack{k+j}}_q .$ $\endgroup$ Apr 25, 2015 at 12:37
  • $\begingroup$ For example for $a(j)= {\frac{j(3j-1)}{2}}$ we get $r(n,k,a(j))= {{n}\brack{k}}_{q}$ and $s(n,k,a(j))= {{n}\brack{k}}_{q}(1+q^{n+1-k}+q^{n+1})$ and for $a(j)= {\binom{j}{2}}$ we have $r(n,k,a(j))= q^{(n+1)k-k^2}{{n}\brack{k}}_{q}$ and $s(n,k,a(j))= q^{nk-k^2}{{n}\brack{k}}_{q}.$ $\endgroup$ Apr 25, 2015 at 12:43


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