Generating Functions with Fibonacci a) Let
\begin{align*}
F_{\text{even}}(x) &= F_0x^0 + F_2x^2 + F_4x^4 + F_6x^6 + F_8x^8 + \cdots \\
&= x^2 + 3x^4 + 8x^6 + 21x^8 + \cdots
\end{align*}
be the generating function whose coefficient of $x^n$ is the $n^{\text{th}}$ Fibonacci number for even $n$, and is zero for odd $n$.
Write $F_{\text{even}}(x)$ as a rational function (that is, as a simplified quotient of polynomials).
b) For what ordered pair of constants $(a,b)$ is it true that $F_{2n}=aF_{2n-2}+bF_{2n-4}$ for all integers $n\ge 2$?

How can I approach this with generating functions?
 A: Repeating Slade's hint, for every generating function $F$, it is the case that $$F_{\mathrm{even}}(x) = \frac{F(x) + F(-x)}{2}. $$
Similarly, we have
$$F_{\mathrm{odd}}(x) = \frac{F(x) - F(-x)}{2}. $$
Even more generally, for any $k$ and $m$ we can consider
$$ F_{m,k} = \frac{1}{m} \sum_{t=0}^m \omega^{kt} F(\omega^{-t} x), $$
where $\omega$ is a primitive $m$th root of unity. The coefficient of $x^n$ in $F_{m,k}$ is
$$
F_n \frac{1}{m} \sum_{t=0}^m \omega^{(k-n)t}.
$$
When $n \equiv k \pmod{m}$, the sum equals $1$, and otherwise it vanishes. So $F_{m,k}$ contains those powers $n$ such that $n \equiv k \pmod{m}$.
As Lucian comments, for your particular case you get
$$
F_{\mathrm{even}}(x) = \frac{x^2}{x^4 - 3x^2 + 1},
$$
which implies that
$$
(x^4 - 3x^2 + 1) F_{\mathrm{even}}(x) = x^2.
$$
You should be able to deduce $a,b$ from this.
A: By making use of 
\begin{align}
F_{n} = \frac{\alpha^{n} - \beta^{n}}{\alpha - \beta}
\end{align}
where $2 \alpha = 1+ \sqrt{5}$ and $2 \beta = 1 - \sqrt{5}$, then
\begin{align}
\sum_{n=0}^{\infty} F_{2n} \, x^{2n} &= \frac{1}{\alpha - \beta} \, \left( \frac{1}{1-\alpha^{2} x^{2}} - \frac{1}{1 - \beta^{2} x^{2}} \right) = \frac{F_{2} \, x^{2}}{1 - L_{2} x^{2} + x^{4}}
\end{align}
where $L_{n}$ are the Lucas numbers. 
By using the difference equation $F_{2n+4} = a F_{2n+2} + b F_{2n}$ then
\begin{align}
\sum_{n=0}^{\infty} F_{2(n+2)} \, x^{2n} &= a \sum_{n=0}^{\infty} F_{2(n+1)} \, x^{2n} + b \sum_{n=0}^{\infty} F_{2n} \, x^{2n} \\
\sum_{n=2}^{\infty} F_{2n} \, x^{2n-4} &= a \sum_{n=1}^{\infty} F_{2n} \, x^{2n-2} + b \sum_{n=0}^{\infty} F_{2n} \, x^{2n} \\ 
\frac{1}{x^{4}} \, \left( - F_{2} x^{2} + \frac{F_{2} x^{2}}{1 - L_{2} x^{2} + x^{4}} \right) &= \frac{a}{x^{2}} \, \frac{F_{2} \, x^{2}}{1 - L_{2} x^{2} + x^{4}} + \frac{b \, F_{2} \, x^{2}}{1 - L_{2} x^{2} + x^{4}} \\
\end{align} 
which becomes $F_{2} L_{2} x^{4} - F_{2} x^{6} = a F_{2} x^{4} + b F_{2} x^{6}$ and leads to $a = L_{2} = 3$ and $b = -1$. This can be verified in the following way:
\begin{align}
3 F_{2n-2} - F_{2n-4} &= 3 F_{2n-2} - ( F_{2n-2} - F_{2n-3} ) \\
&= F_{2n-1} + F_{2n-2} \\
&= F_{2n}
\end{align}
which is the desired result. 
