coefficient of a polynomial 
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.
 A: It might be more helpful to factorize the quadratic expression first (taking it as a variable in $x$). This way we can extract coefficient of $x^n$ ($u$ taken as a constant) and then coefficient of $u^m$ (in other words $[x^n u^m]f(x,u)=[u^m]([x^n]f(x,u))$. So, by factorization of denominator we arrive at
$$
\dfrac{1}{1-2x+x^2-ux^2}=\frac{1}{1-(1+\sqrt{u})x}\cdot \frac{1}{1-(1-\sqrt{u})x}
$$
which by geometric series is
$$
(\sum_{i \geq 0}(1+\sqrt{u})^ix^i) \cdot (\sum_{j \geq 0}(1-\sqrt{u})^j x^j ).
$$
So we get a coefficient of $x^n$
$$
\sum_{k=0}^{n}(1+\sqrt{u})^k(1-\sqrt{u})^{n-k}\tag{*}
$$
and the problem reduces to finding coefficient of $u^m$ in $(*)$. We can evaluate the expression for example by writing it as
$$
(1-\sqrt{u})^n\sum_{k=0}^{n}\left(\frac{1+\sqrt{u}}{1-\sqrt{u}}\right)^k
$$
and spot the finite geometric series with $q=\frac{1+\sqrt{u}}{1-\sqrt{u}}$, so we can just use well-known formula for the sum $\frac{q^{n+1}-1}{q-1}$. After some messy algebra we get
$$
\frac{1}{2\sqrt{u}}[(1+\sqrt{u})^{n+1}-(1-\sqrt{u})^{n+1}],
$$
which finally by Binomial theorem gives
$$
\frac{1}{2\sqrt{u}}\sum_{m=0}^{n+1}\binom{n+1}{m}\sqrt{u}^{m}(1-(-1)^{m}).
$$
For even $m$ the terms vanish and we are left with
$$
\sum_{m=0}^{\lfloor n/2 \rfloor}\binom{n+1}{2m+1}u^{m}.
$$
Now just read off the coefficient, also perhaps use $\binom{n}{k}=\binom{n}{n-k}$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Show that the coefficient of $\ds{\bracks{x^{n}u^{m}}}$ in the bivariate generating function $\ds{1 \over 1 - 2x + x^{2} - ux^{2}}$ is
$\ds{\bbox[5px,#ffd]{n + 1\choose n - 2m}: {\Large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{\bracks{x^{n}u^{m}}{1 \over 1 - 2x + x^{2} - ux^{2}}} =
\bracks{x^{n}u^{m}}{1 \over \pars{1 - x}^{2} - ux^{2}}
\\[5mm] = &\
\bracks{x^{n}u^{m}}{1 \over \pars{1 - x}^{2}}
\bracks{1 - {x^{2} \over \pars{1 - x}^{2}}\,u}^{-1} =
\bracks{x^{n}}{1 \over \pars{1 - x}^{2}}
\bracks{x^{2} \over \pars{1 - x}^{2}}^{m}
\\[5mm] = &\
\bracks{x^{n - 2m}}\pars{1 - x}^{-2m - 2} =
{-2m - 2 \choose n - 2m}\pars{-1}^{n - 2m}
\\[5mm] = &\
{-\bracks{-2m - 2} + \bracks{n - 2m} - 1 \choose n - 2m} =
\bbx{\large{n +  1 \choose n - 2m}} \\ &
\end{align}
