So I have the recurrence relation $c_n = 3 c_{n-1} - c_{n-2}$ and have calculated its generating function to be $F(z) = \frac{1}{1-3z+z^2}$.
I now need to find an explicit expression for $c_n$, but I'm really not sure how to do this, if someone could help me out that would be great.
Thank you very much for your time.
With a bit of experience one may recognize in $x^2-3x+1$ the minimal polynomial of $\varphi^2$ over $\mathbb{Q}$, where $\varphi=\frac{1+\sqrt{5}}{2}$ is the golden ratio. This is enough to be able to state that $[z^n]\frac{1}{1-3z+z^2}$ only depends on $F_{2n}$ and $F_{2n+2}$. Actually
$$ \sum_{n\geq 0} F_n z^n = \frac{z}{1-z-z^2} $$ implies $$ \sum_{n\geq 0} F_{2n} z^{2n} = \frac{1}{2}\left[ \frac{z}{1-z-z^2}+ \frac{-z}{1+z-z^2}\right] = \frac{z^2}{1-3z^2+z^4}$$ and $$ \sum_{n\geq 0} F_{2n}\,z^{n} = \frac{z}{1-3z+z^2}$$ so $$ \frac{1}{1-3z+z^2} = \sum_{n\geq 0}\color{red}{F_{2n+2}}\,z^n.$$