Take $q$ a primitive $r$-root of unit, that is $q^r=1$ and $q^k \neq 1$, for all $k<r$. Consider the following expression defined by double recursive formula: $$ F(l,n) = q^{-l}F(l,n-1) - q^{-l}F(l+1,n-1) \ \ \textrm{ and } \ \ F(l,0)=\delta_{l,0} $$ we also assume that $F(l+r,n) = F(l,n)$.
Let $\displaystyle {n \choose k}_q$ by the $q$-binomial number, which can be defined also recursively by $$ {n \choose k}_q = q^k{n-1 \choose k}_q + {n-1 \choose k-1}_q $$ also, $$ {0 \choose 0}_q = 1, \ \ \ \ {n \choose k}_q=0, \textrm{ if } k>n. $$ Now that my question is if the following formula is satisfied: \begin{eqnarray} \sum_{k=0}^{n}q^{t(n-k)}{n \choose k}_qF(l+t, n+m-k)F(l+m-k,k)= \\ = \sum_{k=0}^{n}q^{tk}{n \choose k}_qF(l, n-k)F(l+m+t-k,m+k). \end{eqnarray} I already know that it works for $r=2,3$ by hand.