<s>[INST] Solving recurrence equation using generating function method I've been trying to solve the following equation intuitively (I only know the method if there are minuses in the equation - $a_{n-1}, a_{n-2}...$).
$$a_{n+2}=4a_{n+1}-4a_{n}$$
$$a_{0}=3$$
$$a_{1}=8$$
$$
\begin{align}
A(x)&=\sum\limits_{n>=0}a_{n}x^{n} \\
&= \sum\limits_{n>=0}(a_{n+1}-\frac{1}{4}a_{n+2})x^{n} \\
&= \sum\limits_{n>=0}(a_{n+1})x^{n}-\frac{1}{4}\sum\limits_{n>=0}(a_{n+2})x^{n} \\
&= \sum\limits_{n>=1}(a_{n})x^{n+1}-\frac{1}{4}\sum\limits_{n>=2}(a_{n})x^{n+2} \\
&= \frac{1}{x}\sum\limits_{n>=1}(a_{n})x^{n}-\frac{1}{4x^{2}}\sum\limits_{n>=2}(a_{n})x^{n} \\
&= \frac{1}{x}[\sum\limits_{n>=0}(a_{n})x^{n} - 3]-\frac{1}{4x^{2}}[\sum\limits_{n>=0}(a_{n})x^{n} - 3 - 8x] \\
&= \frac{1}{x}[A(x) - 3]-\frac{1}{4x^{2}}[A(x) - 3 - 8x]
\end{align}
$$
So I get
$$A(x)=\frac{-4x+3}{4x^{2}-4x+1}$$
Is this correct? I'm asking because the answer to this question according to the source I got it from is $x^{2}-4x+4$ as the denominator...
 [/INST] Not sure where you went wrong, so I'll start from scratch with a different method:
Let $A(x)=\sum_{n=0}^\infty a_nx^n$.  Note that $a_n$ satisfies $a_{n+2}-4a_{n+1}+4a_n=0$.  Let $p(x)=1-4x+4x^2$.  We then define
$$
B(x) = \sum_{n=0}^\infty b_nx^n=p(x)A(x)
$$
Note that for $n\geq2$, we have $b_n=a_{n+2}-4a_{n+1}+4a_n=0$. So, we simply have
$$
p(x)A(x) = b_0+b_1 x
$$
Thus,
$$
A(x) = \frac{b_0+b_1 x}{1-4x+4x^2}=\frac{a_0+(a_1-4a_0) x}{1-4x+4x^2}=\frac{3-4x}{1-4x+4x^2}
$$
It seems you have the right answer, and that your source has a typo.
</s>