Is there a $q$-analogue of the formula $\sum_{k=0}^n \, {n \choose k} = 2^n $ in terms of the $q$-binomial coefficient ${n \choose k}_q$ and $(2^n)_q=(1+q)...(1+q^n)$?
2025-01-13 05:52:42.1736747562
$q$-analogue of $\sum_{k=0}^n \, {n \choose k} = 2^n $
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The most natural analog is $\sum_{k=0}^{n}q^{k}\binom{n}{k}_{q^2}=(1+q)(1+q^2)\dots(1+q^n)$. If you can read German you find a proof in https://homepage.univie.ac.at/johann.cigler/preprints/id2.pdf. (2.16).