$q$-binomial sum, slightly Recall that $[n]_{q}!=\prod_{j=1}^n\frac{1-q^{j}}{1-q}$ and $\binom{n}k_{q}=\frac{[n]_{q}!}{[k]_{q}![n-k]_{q}!}$. Then the $q$-binomial theorem states
$$\sum_{k=0}^n\binom{n}k_qq^{\binom{k}2}=\prod_{k=0}^{n-1}(1+q^k).$$
I wanted to make a slight change to this and ask:

QUESTION. Is there a formula for the following? What is its combinatorial interpretation, if any?
$$\sum_{k=0}^n\binom{n}k_{q^2}q^{3k}.$$

 A: Let $f_{n,a}(q)=\sum_{k=0}^n\binom{n}k_{q^2}q^{ak}$ and $(-q;q)_n=(1+q)(1+q^2)\cdots(1+q^n)$.
The familiar recurrence
$\binom{n+1}k_{q^2}=\binom{n}k_{q^2}+q^{2n+2-2k}\binom{n}{k-1}_{q^2}$, the symmetry $\binom{n}k_{q^2}=\binom{n}{n-k}_{q^2}$ followed by the replacement $k\rightarrow n-k$ (in the sum) give out
\begin{align} f_{n+1,1}(q)&=\sum_{k=0}^{n+1}\binom{n}k_{q^2}q^k+\binom{n}{k-1}_{q^2}q^{2n+2-k}
=\sum_{k=0}^n\binom{n}k_{q^2}q^k+q^{n+1}\sum_{k=0}^n\binom{n}k_{q^2}q^{n-k} \\
&=f_{n,1}(q)+q^{n+1}\sum_{k=0}^n\binom{n}{n-k}_{q^2}q^{n-k}=(1+q^{n+1})f_{n,1}(q).
\end{align}
We arrive at $f_{n,1}(q)=(-q;q)_n$, inductively. Let's turn to $f_{n,3}(q)$ for which one invokes the recurrence
$\binom{n+1}k_{q^2}=q^{2k}\binom{n}k_{q^2}+\binom{n}{k-1}_{q^2}$. Proceed the same as above to obtain
\begin{align} 
f_{n+1,1}(q)&=\sum_{k=0}^{n+1}\binom{n}k_{q^2}q^{3k}+\binom{n}{k-1}_{q^2}q^k
=\sum_{k=0}^n\binom{n}k_{q^2}q^{3k}+q\sum_{k=0}^n\binom{n}k_{q^2}q^k \\
&=f_{n,3}(q)+qf_{n,1}(q).
\end{align}
That means $f_{n,3}(q)=(-q;q)_{n+1}-q(-q;q)_n$ answers my question.
Motivated by Richard Stanley's comments, one may emulate our argument to compute $f_{n,a}(q)$ for $a$ odd. Here is the assertion: denote the elementary symmetric polynomials by $e_0=1$ and $e_j(x_1,\dots,x_m)=\sum_{1\leq i_1<i_2<\cdots<i_j\leq m}x_{i_1}\cdots x_{i_j}$ for $1\leq j\leq m$. Then,
$$f_{n,2a+1}(q)=\sum_{j=0}^{a-1}(-1)^j\,e_j(q^3,q^5,\dots,q^{2a-1})\cdot h_{n+a-j-1}(q)$$
where $h_n(q):=(-q;q)_{n+1}-q(-q;q)_n=f_{n,3}(q)$.
I would like to leave the proof to the interested reader.
