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?
Best Answer
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.