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.



Quest for a human proof of a $q-$binomial identity
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.


