Show that the coefficient of $[x^nu^m] $ in the bivariate generating function $\dfrac{1}{1-2x+x^2-ux^2}$ is ${n+1\choose n-2m}.$
I tried to do this by using the multinomial theorem (an extension of the binomial theorem), which basically states that for terms $x_1,\cdots, x_r, n\in \mathbb{N}_{\geq 0}, (x_1+\cdots + x_r)^n = \sum_{k_1+\cdots + k_r = n} \dfrac{n!}{k_1! \cdots k_r!}x_1^{k_1}\cdots x_r^{k_r}.$
This gives that the given bivariate generating function is equal to $\sum_{n\geq 0}(2x-x^2+ux^2)^n = \sum_{n\geq 0} \sum_{k_1+k_2 + k_3 = n} \dfrac{n!}{k_1!k_2!k_3!} (2x)^{k_1}(-x^2)^{k_2}(ux^2)^{k_3}$.
Thus the coefficient of $[x^n u^m]$ should be $\sum_{k_1 + 2k_2 = n-2m} \dfrac{(n-k_2-m)!}{k_1!k_2!m!}2^{k_1} (-1)^{k_2} .$ I can further simplify this by replacing $k_2$ with $\dfrac{n-2m-k_1}{2},$ but I'm not sure how to get the desired result from that. Is there some other useful property of polynomials? I also realized that $\sum_{m\geq 0} {n+1\choose n-2m} = 2^n,$ which can be shown using Pascal's identity, though I'm not sure if this is useful.