Express the sequence $\{x_n\}$ where $x_n = 2x_{n-1} + 3x_{n-2}$ in terms of ${x_0, x_1, n}$ Given a sequence $\{x_n\}$ where $x_n = 2x_{n-1} + 3x_{n-2}$ how can one express it in terms of ${x_0, x_1, n}$. Can this be generalized for ${x_n = \alpha x_{n-1} + \beta x_{n-2}}$
I've tried to use the following approach:
$$\eqalign{
 & 1x_0 = x_01 \\
 & zx_1 = x_1z \\
 & z^2x_2 = (2x_1 + 3x_0)z^2 \\
 & z^3x_3 = (2x_2 + 3x_1)z^3 \\
 & z^4x_4 = (2x_3 + 3x_2)z^4 \\
 & ... \\
 & z^nx_n = (2x_{n-1} + 3x_{n-2})z^n
}$$
Then sum LHS with RHS which will produce:
$${
x_0 + \sum\limits_{k = 1 }^na_nz^n = x_0 +zx_1 + 2\sum\limits_{k = 2}^nx_{n-1}z^n + 3 \sum\limits_{k = 2}^n x_{n-2}z^n
}$$
Let 
$${
G(z) = x_0 + \sum\limits_{k = 1 }^na_nz^n
}$$
Then RHS may be expressed in terms of ${G(z)}$. For example 
$${
2\sum\limits_{k = 2}^na_{n-1}z^n = 2z\sum\limits_{k = 2}^na_{n-1}z^{n-1} = 2z(\sum\limits_{k = 1}^na_{n}z^{n} + x_0 - x_0) = 2z(G(z) - x_0)
}$$
Applying those transformations I eventually got ${G(z)}$ expressed in terms of z and ${x_1, x_0}$. But at this point I got stuck.
I got a sum of fractions:
$${
\frac{3x_0 - x_1}{4(1+z)}+\frac{x_0+x_1}{4(1-3z)}
}$$
I guess i could expand the fractions into series and find their sum, but i am not supposed to know about such expansions at the point of the book i took the problem from.
All of the above feels like a wrong approach. So the question is whether this can be done in a more elegant way.

Let $f(x)$ be the generating function for some sequence $(a_{n})_{n\in\mathbb{N}}$ such that 
    \begin{align*}
  f(x)=\frac{c}{bx+d}=\sum^{\infty}_{n=0}{a_{n}x^{n}},\ c,b,d\in\mathbb{Z}.\tag{1}
 \end{align*} 
Then
\begin{alignat*}{3}
  f(x)=\frac{c}{bx+d}&\implies f^{\prime}(x)                       &&=-1!\left(\frac{b}{c}\right)^{1}f(x)^{2}\\
                               &\implies f^{\prime\prime}(x)             &&=+2!\left(\frac{b}{c}\right)^{2}f(x)^{3}\\
                               &\implies f^{\prime\prime\prime}(x)   &&=-3!\left(\frac{b}{c}\right)^{3}f(x)^{4}\\
                               &\qquad                                            &&\vdots\\
                               &\implies f^{n}(x)                               &&=(-1)^{n}n!\left(\frac{b}{c}\right)^{n}\big(f(x)\big)^{n+1},
 \end{alignat*}
where $f^{n}(x)$ denotes the $n^{\text{th}}$ derivative of $f$ with respect to $x.$ If we evaluate $f^{n}(0)$, the Maclaurin series expansion of $f(x)$ is then given by
\begin{alignat*}{4}
   &\qquad \qquad \qquad \qquad \quad f^{n}(0)&&=(-1)^{n}n!\left(\frac{b}{c}\right)^{n}f(0)^{n+1}\\
   &              &&=(-1)^{n}n!\left(\frac{b}{c}\right)^{n}\left(\frac{c}{d}\right)^{n+1}\\
   &\qquad \qquad \qquad \quad \implies f(x)        &&=\sum_{n=0}^{\infty }c\ (-1)^{n}\left(\frac{1}{d}\right)^{n+1}{b^{n}x^{n}}.\tag{2}\\
     \end{alignat*}
It follows from (1) and (2) that the explicit form of the sequence $(a_{n})_{n\in\mathbb{N}}$ is given by
 \begin{align*}a_{n}=c\ (-1)^{n}\left(\frac{1}{d}\right)^{n+1} b^{n}.\tag{3}\\ \end{align*}
If we let $c=x_{0}+x_{1},\ b=-12,$ and $d=4$, as in (1) then 
\begin{alignat*}{3}
  g(x)=\frac{x_{0}+x_{1}}{4-12z}&\overset{(3)}{\implies} g_{n}&&=(-1)^{n}(x_{0}+x_{1})(-12)^{n}\left(\frac{1}{4}\right)^{n+1}\\
                               &\qquad              &&=(x_{0}+x_{1})(-1)^{n}(-1)^{n}(3)^{n}(4)^{n}\left(\frac{1}{4}\right)^{n}\left(\frac{1}{4}\right)\\
                               &\ \implies g_{n}          &&=\left(\frac{1}{4}\right)3^{n}(x_{0}+x_{1}).\tag{4}
 \end{alignat*}
Similarly
\begin{alignat*}{3}
  h(x)=\frac{3x_{0}-x_{1}}{4+4z}&\overset{3}{\implies} h_{n}&&=(-1)^{n}(3x_{0}-x_{1})(4)^{n}\left(\frac{1}{4}\right)^{n+1}\\
                               &\qquad              &&=(3x_{0}-x_{1})(-1)^{n}(4)^{n}\left(\frac{1}{4}\right)^{n}\left(\frac{1}{4}\right)\\
                               &\ \implies h_{n}          &&=(-1)^{n}\left(\frac{1}{4}\right)(3x_{0}-x_{1}).\tag{5}\\
 \end{alignat*}
It follows from the linearity of differentiation, (superposition principle), that we may combine (4) and (5) to obtain the desired formula for $x_{n}$. Whence,
\begin{align}x_{n}=\left(\frac{1}{4}\right)3^{n}(x_{0}+x_{1})+ (-1)^{n}\left(\frac{1}{4}\right)(3x_{0}-x_{1})\tag*{$\square$}\end{align}
