We all know that $\sum_{i=0}^{n}{n \choose i}=2^{n}$. Is there a similar result regarding the qbinomial coefficients? (a.k.a Gaussian binomial coefficients)  $\sum_{i=0}^{n}{n \choose i}_{q}=?$

The identity $\prod_{i=0}^{n1} (1+xq^i) = \sum_{k=0}^n x^k q^{{k\choose 2}}{n\choose k}_q$ is the $q$binomial theorem. A combinatorial proof based on integer partitions is mentioned on page 68 of Enumerative Combinatorics, vol. 1, 2nd ed. There is also given a combinatorial proof based on finite fields. For the online version at http://math.mit.edu/~rstan/ec/ec1.pdf, see pages 7475. 


There are many possibilities, e.g. $\sum_{i=0}^{n}q^i{n \choose i}_{q^2}=(1+q)(1+q^2)...(1+q^n)$ or $\sum_{i=0}^{n}q^{i(i+1)/2}{n \choose i}_{q }=(1+q)(1+q^2)...(1+q^n).$ 


For Gaussian binomial coefficients we have $$ \sum_{k = 0}^n \binom nk_q = \sum_{m = 0}^\infty a_m q^m, $$ where $$ a_m = \sum_{\lambda\vdash m} \#\{k\in \mathbf Z_{\geq 0}\mid \lambda_1\leq nk, \lambda'_1\leq k\}. $$ The notation $\lambda\vdash m$ signifies that $\lambda$ is an integer partition of $m$. Also $\lambda_1$ is the first (largest) part of $\lambda$ and $\lambda'_1$ is the number of positive parts in $\lambda$. This follows from the following wellknown fact about Gaussian binomial coefficients: $$ \binom nk_q = \sum_\lambda q^{\lambda}, $$ the sum being over all partitions $\lambda$ where $\lambda_1\leq nk$ and $\lambda'_1\leq k$ (see, for example Stanley's Enumerative Combinatorics, vol. 1, or Eq. (4) in my expository article titled Counting subspaces in a Finite Vector Space). 


One more version  analog of $\sum_{i=0}^n(1)^i\binom ni=0$: $$ \sum_{i=0}^n(1)^i\binom ni_q=\begin{cases}0,&n=2k1\\ \prod_{j=1}^k(1q^{2j1}),&n=2k\end{cases} $$ 

