Let $[x]\_q=\frac{1-q^x}{1-q}$, $[n]\_q!=[1]\_q[2]\_q\cdots[n]\_q$ and ${\binom{x}{n}}\_{q}=\frac{[x]\_q[x-1]\_q\cdots[x-n+1]\_q }{[n]\_q!}$.


Computer experiments suggest that
$$\det \left(q^\binom{i-j}{2}\left(\binom{i+r}{j}\_{q}x+\binom{i+r-j}{j}\_{q}\right)\right)\_{i,j = 0}^{n - 1} = (1+x)^n$$


Any idea how to prove this?



A q-rious identity
Let $A\_n$ be the matrix involved in the problem and let $L\_n=\left((-1)^{i-j}\binom{i}{j}\_q\right)\_{i,j=0}^{n-1}$. 
Observe that $L\_n$ is lower-triangular with 1's in the diagonal. Multiplying, we have: 
$$L\_nA\_n=\left(x\,u\_{i,j}(r)+u\_{i,j}(r-j)\right)\_{i,j=0}^{n-1}$$
where 
$$u\_{i,j}(r)=\sum\_{k=0}^i(-1)^{i-k}q^{\binom{k-j}{2}}\binom{i}{k}\_q\,\binom{k+r}{j}\_q\,.$$


Now
\begin{eqnarray\*}
u\_{i,j}(r)&=&\sum\_{k=0}^i(-1)^{i-k}q^{\binom{k-j}{2}}\binom{i}{i-k}\_q\,\binom{k+r}{k+r-j}\_q \\
&=&(-1)^{r+i-j}\,q^{\binom{r+1}{2}}\,\sum\_{k=0}^i q^{k(k+r-j)}\binom{i}{i-k}\_q\,\binom{-j-1}{k+r-j}\_q \\
&=&(-1)^{r+i-j}\,q^{\binom{r+1}{2}}\,\sum\_{k=0}^i q^{k(k+r-j)}\binom{i}{i-k}\_q\,\binom{-j-1}{k+r-j}\_q \\
&=&(-1)^{r+i-j}\,q^{\binom{r+1}{2}}\,\binom{i-j-1}{i-j+r}\_q
=q^{\binom{i-j}{2}}\,\binom{r}{j-i}\_q\,.
\end{eqnarray\*}
Hence $u\_{i,j}(r)=0$ when $i>j$ and $u\_{i,i}(r)=1$.
Thus $L\_nA\_n$ is upper triangular with $x+1$ as elements of the diagonal.


