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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.

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2 Answers 2

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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.

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  • $\begingroup$ Dear @Johann, I added an answer below. Could you please check it ? Thank you. $\endgroup$
    – drxy
    Apr 29, 2020 at 19:48
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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 $$ \sum_{k=0}^{n} \left( \left[\begin{array}{l} n \\ k \end{array}\right]_{q}q^{k \choose 2} \right)^2, \quad \sum_{k=0}^{n} \left( \left[\begin{array}{l} n \\ k \end{array}\right]_{q} \right)^2. $$

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    $\begingroup$ $\sum_{k}{\binom{n}{k}}_{q} q^{\frac{k^2}{2}}$ $=(-q^{\frac{1}{2}} ;q)_n $. $\endgroup$ Apr 30, 2020 at 9:29
  • $\begingroup$ @JohannCigler Thank you very much. Meanwhile, is there any way to represent the sums $$\sum_{k=0}^{n} \left( \left[\begin{array}{l} n \\ k \end{array}\right]_{q}q^{k \choose 2} \right)^2, \quad \sum_{k=0}^{n} \left( \left[\begin{array}{l} n \\ k \end{array}\right]_{q} \right)^2$$ via q-Pochhammer symbol ? $\endgroup$
    – drxy
    Apr 30, 2020 at 10:01
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    $\begingroup$ No. You need only compute these sums for small values of $n$. There is no factorization into terms $1-q^j x$ for some $x.$ $\endgroup$ Apr 30, 2020 at 11:07
  • $\begingroup$ @JohannCigler thank you very much for your great efforts. I'm accepting your solution. If you find any results, you can update your answer. I will follow this topic. $\endgroup$
    – drxy
    Apr 30, 2020 at 11:21

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