# Combinatorial Identity $\sum_{k=1}^n (-1)^{k-1} \cdot q^{\frac{k(k-1)}{2}} \cdot \frac{\prod_{i=n-k+1}^n(1-q^i)}{\prod_{i=1}^k(1-q^i)} = 1$

I have to validate the following identity which is defined: $$\sum_{k=1}^n (-1)^{k-1} \cdot q^{\frac{k(k-1)}{2}} \cdot \frac{\prod_{i=n-k+1}^n(1-q^i)}{\prod_{i=1}^k(1-q^i)} = 1$$ where $0<q<1$.

I verified the expression numerically with R/Matlab by choosing different values for $q$ and so far it is correct. I also verified trivial cases where $n=2,3,4$ and it is also correct. But I couldn't figure out how one can prove it for all of $n$.

I hope someone could give me a hint to proceed or share some experiences in proving this kind of identities. Thanks in advance!

Edit: I thought giving a context might be interesting. The original identity that I am working is: $$\sum_{k=1}^n \frac{1}{(1-q^k)(q^{-1},q^{-1})_{k-1}(q,q)_{n-k}} = \frac{1}{(q^{-1},q^{-1})_0 (q,q)_n} = \frac{1}{(q,q)_n}$$ where $(q,q)_n = \prod_{i=1}^n (1-q^i)$ is the q_Pochhamer symbol with $(q,q)_0 = 1$. These two identities are equivalent, I need to prove at least one :P

Update: I have demonstrated the following identity, in case of somebody is interested: $$\sum_{k=1}^n \frac{1}{(q^{-1},q^{-1})_{k-1}(q,q)_{n-k}} = 1$$

Even though this and the original are surprisingly close to each other, I couldn't find away to demonstrate the first identity.

• By induction maybe? – Avitus Dec 11 '13 at 14:12
• I did try induction but it didn't lead to anywhere. As I see, the $n$ step doesn't bring any useful information for the $n+1$ step. Nevertheless, I keep digging the induction way 'cause it's the only 'obvious' method for the time being. Thanks anyway! – Tuan Dec 11 '13 at 14:35

You're asking why $$\sum(-1)^kq^{k(k-1)/2}\genfrac[]{0pt}{}nk=0.$$ That's a $q$-deformed version of the classical identity $\sum(-1)^k\binom nk=0$ (which corresponds to $q=1$). At least some proofs of the classical identity can be generalized to the «quantum» case. For instance, there is (as I've mentioned in comments) $q$-binomial theorem (that can be prove either purely combinatorially or by induction), $$(1+x)(1+xq)(1+xq^2)\ldots(1+xq^{n-1})=\sum x^kq^{k(k-1)/2}\genfrac[]{0pt}{}nk$$ — now plug $x=-1$.