How to find some specific sums of $q$-binomial coefficients? We know that sums of binomial coefficients are
$$
 \sum_{k=0}^{n}{\binom{n}{k}^2}=\binom{2n}{n} \quad \text{and} \quad  \sum_{k=0}^{n}{\binom{n}{k}}=2^n.
$$
First equality can be proven via Vandermonde identity by setting $m=r=n$ as:
$$
{m+n \choose r} = \sum_{k=0}^r {m\choose k}{n\choose r-k}.
$$
Now, I want to find various sums of the $q$-binomial coefficients. Thus, how can I find the following sums by using $q$-binomial properties? 
$$
\sum_{k=0}^{n}
\left( 
\left[\begin{array}{l}
n \\
k
\end{array}\right]_{q}q^{k \choose 2} 
\right)^2, \quad 
\sum_{k=0}^{n}
\left( 
\left[\begin{array}{l}
n \\
k
\end{array}\right]_{q}
\right)^2,\quad
\sum_{k=0}^{n}
\left[\begin{array}{l}
n \\
k
\end{array}\right]_{q} \quad \text{and} \quad 
\sum_{k=0}^{n}
\left[\begin{array}{l}
n \\
k
\end{array}\right]_{q}q^{\frac{k^2}{2}},
$$
where $\left[\begin{array}{c}
m \\
r
\end{array}\right]_{q}=\frac{[n]_{q} !}{[k]_{q} ![n-k]_{q} !} \quad(k \leq n)$ 
and  $[n]_{q}= \frac{1-q^n}{1-q}$.
I have tried to proof via $q-$Vandermonde matrix but I couldn't achieve. 
 A: From the well-known formula $$\sum_{k}{\binom{n}{k}_{q} q^{\binom{k}{2}}x^k}=(1+x)(1+qx)\dots(1+q^{n-1}x)$$ you get a formula for  $\sum_{k}{\binom{n}{k}_{q}}q^{k^2/2}$.
For the other sums you only get recurrences, for example with the q-Zeilberger algorithm (cf. https://risc.jku.at/sw/qzeil/).
Natural $q-$analogues of your sums are
$$\sum_{k}{q^{\binom{k+1}{2}}}{\binom{n}{k}}_{q}=
\sum_{k} q^{k} \binom{n}{k}_{q^2}=(1+q)(1+q^2)\dots (1+q^n)$$
and
$$\sum_{k}{q^{k^2}}\binom{n}{k}_{q}^2=\binom{2n}{n}_q.$$
Edit
Let $s(n,q)=\sum_{k} \binom{n}{k}_{q}.$  There is no closed formula, but we get the recursion $$s(n,q)=2s(n-1,q)+(q^{n-1}-1)s(n-2,q),$$ which for $q=1$  reduces to $s(n,1)=2s(n-1,1).$
Let $t(n,q)=\sum_{k}\binom{n}{k}_{q}^2.$
Then we get
 $$t(n,q)=\frac{2+q-q^{2n-1}-2q^n}{1-q^n}t(n-1,q)-\frac{(1-q^{n-1})^2(1+2q+q^n)}{1-q^n}t(n-2,q)+\frac{q(1-q^{n-1})^2(1-q^{n-2})^2}{1-q^n} t(n-3,q).$$
For $q\rightarrow 1$ we get $t(n,1)=(2+\frac{2n-2}{n}t(n-1,1)=\frac{2n(2n-1)}{n^2}t(n-1,1),$
which gives $t(n,1)=\binom{2n}{n}.$
For the third sum we get a similar, but more complicated, recursion.
A: Thank you very much for your notable answer. According to your formula given above,  can we find the sums expressed by q-Pochhammer symbol as 
$$
\sum_{k}{\binom{n}{k}_{q}}q^{k^2/2} \stackrel{?}{=} (-q^{\frac{k}{2}};q)_n \quad \text{and} \quad \sum_{k}{\binom{n}{k}_{q}} \stackrel{?}{=} (-q^{-\binom{n}{2}};q)_n .
$$
On the other hand, by $q-$Vandermonde identity 
$$\left(\begin{array}{c}m+n \\ k\end{array}\right)_{q}=\sum_{j}\left(\begin{array}{c}m \\ k-j\end{array}\right)_{q}\left(\begin{array}{l}n \\ j\end{array}\right) q^{j(m-k+j)},
$$
your second result is clear. How can we find another sums 
$$
\sum_{k=0}^{n}
\left( 
\left[\begin{array}{l}
n \\
k
\end{array}\right]_{q}q^{k \choose 2} 
\right)^2, \quad 
\sum_{k=0}^{n}
\left( 
\left[\begin{array}{l}
n \\
k
\end{array}\right]_{q}
\right)^2.
$$
