Timeline for How to prove that $\sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i}=\frac{(ab;q)_n}{(q;q)_n}$?
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| Oct 26, 2016 at 5:34 | comment | added | Ofir Gorodetsky | I see. By plugging negative powers of $q$, instead of positive ones as I did, one gets a much simpler proof. Very nice! | |
| Oct 25, 2016 at 22:57 | history | answered | Fedor Petrov | CC BY-SA 3.0 |