Show that for all $a\geq b$ $$\left(\sum_{i=0}^ax^{a-2i}\right)\left(\sum_{i=0}^b x^{b-2i}\right)=\sum_{k=0}^a\sum_{l=0}^{a+b-2k}x^{a+b-2k-2l}$$ where $a,b$ are positive integers.
I tried to prove this using double induction $T(a,b)$ where I show that
- $T(1,1)$ is true.
- If $T(a,1)$ is true, then $T(a+1,1)$ is true.
- If $T(a,b)$ is true, then $T(a,b+1)$ is true.
I face pretty heavy computation starting step 2. Is there a nice faster way proving this identity?
Use the formula for the sum of a finite geometric series $$ \sum_{i=0}^n r^i = \frac{r^{n+1}-1}{r-1} $$ to simplify both sides to $$ \frac{x^{a+b+4} - x^{a-b+2} - x^{-a+b+2} + x^{-a-b}}{(x^2 - 1)^2}. $$ Note that to put the sums into the right form, you'll have to pull out some common factors first; for example in the sum as $i$ goes from $0$ to $a$ on the left, you want to factor out $x^a$, then take $r=x^{-2}$.
As Rob Pratt points out in the comments, the formula doesn't work when $r=1$ (in other words, when $x = \pm 1$). In that case, every term of every sum is $\pm 1$ (depending on the parity of $a$ and $b$) and we should get $(-1)^{a+b}(a+1)(b+1)$ on both sides. We could probably avoid considering this case by making some argument about polynomials being defined by finitely many values...
We should also assume that $x\ne 0$ or else we are dividing by $0$ on both sides.