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.