两个等比数列之和的乘积展开式

hbghlyj

$$(t^m + t^{m-2} + \cdots + t^{-m})(t^n + t^{n-2} + \cdots + t^{-n})= \sum_{k=0}^{\min(m,n)} (t^{m+n-2k} + t^{(m+n-2k)-2} + \cdots + t^{-(m+n-2k)})$$
gr.inc/question/lets-carefully-prove-the-cleb … n-for-su2-represent/

hbghlyj

$$\LHS= \frac{t^{m+2}-t^{-m}}{t^2-1} \cdot \frac{t^{n+2}-t^{-n}}{t^2-1}= \frac{t^{m+n+4} - t^{m-n+2} - t^{-m+n+2} + t^{-m-n}}{(t^2-1)^2}$$
Let $M = \min(m,n)$.
$$\RHS= \frac{1}{t^2-1} \left( \sum_{k=0}^{M} t^{m+n-2k+2} - \sum_{k=0}^{M} t^{-(m+n-2k)} \right)=\frac{t^{m+n+4}-t^{m+n-2M+2} - t^{-m-n+2M+2}+t^{-m-n}}{(t^2-1)^2}$$
LHS = RHS simplifies to:
$$t^{m-n+2} + t^{-m+n+2} = t^{m+n-2M+2} + t^{-m-n+2M+2}$$