It's been awhile since I have done this. The sequence is $\displaystyle a_n = a_{n-1} + 5~a_{n-2}$ with $a_{0}=0$ and $a_{1}=1$.
I found the generating function to be $\displaystyle G(x) = \frac{x}{(1-x-5x^2)}$ but I am lost on where to go from here.
On
The Characteristic equation of the Recurrence relation is $r^2-r-5=0$
So, $a_n=A\cdot r_1^n+B\cdot r_2^n$ where $r-1,r_2$ are the roots of the above Quadratic Equation
So, $r_1,r_2$ are $\frac{1\pm\sqrt{1^2-4\cdot1(-5)}}{2\cdot1}=\frac{1\pm\sqrt{21}}2$
$0=a_0=A+B\implies B=-A\implies a_n=A(r_1^n-r_2^n)$
$1=a_1=A(r_1-r_2)\implies A=\frac1{(r_1-r_2)}=\sqrt{21}$
$\implies a_n=\sqrt{21}\left(\left(\frac{1+\sqrt{21}}2\right)^n-\left(\frac{1-\sqrt{21}}2\right)^n\right)$
On
Linear algebra also gives an answer: writing the recurrence as
$\begin{pmatrix}a_n \\ a_{n-1}\end{pmatrix}=\begin{pmatrix}1 & 5 \\ 1 & 0\end{pmatrix}\begin{pmatrix}a_{n-1} \\ a_{n-2}\end{pmatrix}$,
and using spectral decomposition to find powers of the matrix. Note, that the characteristic polynomial of the matrix equals
$\rho(\lambda)=\lambda^2-\lambda -5$.
Factor the denominator. Slightly unpleasant, since the roots are not rational, but doable. Then use partial fractions to express the generating function in the form $\frac{A}{1-ax}+\frac{B}{1-bx}$.
It is easy to write down the power series for $\frac{1}{1-ax}$ and $\frac{1}{1-bx}$, by using the fact that $\frac{1}{1-t}=1+t+t^2+t^3+\cdots$.
Then you can get an explicit expression for the coefficient of $x^n$ in each of the terms, and hence in the sum.
If you run into trouble, I can give more detail. I got the following partial fractions representation: $$\frac{1}{\sqrt{21}}\left(\frac{1}{1-\frac{1}{2}(1+\sqrt{21})x}-\frac{1}{1-\frac{1}{2}(1-\sqrt{21})x} \right) .$$
Remark: There are (to me) easier ways for this particular problem than the generating functions approach. Characteristic equations, though they involve roughly similar manipulations, feel easier.