<s>[INST] Use the method of generating series to solve the recurrence: $b_n=2b_{n-1}-b_{n-2}$ with $b_0=0, b_1=5$ This is my attempt:

Let
\begin{equation}
    B(x):=\sum^{\infty}_{n=0}b_nx^n.
\end{equation}
Multiplying our recurrence through by $x^n$ and then summing from $2$ to $\infty$
\begin{equation}
    \sum^{\infty}_{n=2}b_nx^n=2\sum^{\infty}_{n=2}b_{n-1}x^n-\sum^{\infty}_{n=2}b_{n-2}x^n.
\end{equation}
Reindexing/shifting the appropriate sums
\begin{equation}
    \sum^{\infty}_{n=0}b_nx^n-b_0-b_1x=2x\sum^{\infty}_{n=0}b_nx^n-2xb_0-x^2\sum^{\infty}_{n=0}b_nx^n.
\end{equation}
Substituting $b_0$ and $b_1$ values
\begin{equation}
    \sum^{\infty}_{n=0}b_nx^n-5x=2x\sum^{\infty}_{n=0}b_nx^n-x^2\sum^{\infty}_{n=0}b_nx^n.
\end{equation}
Replacing the sums with $B(x)$
\begin{equation}
    B(x)-5x=2xB(x)-x^2B(x),
\end{equation}
or equivalently
\begin{equation}
    B(x)=\frac{5x}{1-2x+x^2}=\frac{5x}{(x-1)^2}.
\end{equation}
Decomposing into partial fractions
\begin{equation}
    \frac{5x}{(x-1)^2}=\frac{p_1}{x-1}+\frac{p_2}{(x-1)^2},
\end{equation}
or equivalently
\begin{equation}
    5x=p_1(x-1)+p_2.
\end{equation}
Letting $x=1$ gives us $p_2=5$ and letting $x=0$ gives
\begin{equation}
    0=-p_1+5 \Rightarrow p_1=5.
\end{equation}
So we have
\begin{equation}
    B(x)=\frac{5}{x-1}+\frac{5}{(x-1)^2}.
\end{equation}
Using the generalised binomial theorem to find $b_n$
\begin{equation}
    5(x-1)^{-1}=5(1-x)^{-1}=-5\sum^{\infty}_{n=0}x^n.
\end{equation}
\begin{equation}
    5(x-1)^{-2}=5(1-x)^{-2}=5\sum^{\infty}_{n=0}{-2\choose{n}}(-1)^nx^n=5\sum^{\infty}_{n=0}(n+1)x^n. \tag1
\end{equation}
We arrive at (1) due to
\begin{equation}
    {-2\choose{n}}=\frac{\prod^{n-1}_{i=0}(-2-i)}{n!}=\frac{(-2)...(-n)(-1-n)}{n!}=\frac{(-1)^{n}n!(n+1)}{n!}=(-1)^{n}(n+1).
\end{equation}
Thus, the solution to the recurrence is
\begin{equation}
    b_n=-5+5(n+1)=5n.
\end{equation}

However, this of course isn't right since $b_1$ should be $5$ and not $-15$. So where have I gone wrong here? I feel as though it was during that ${-2}\choose{n}$ step however I can't quite see what the mistake was exactly.

Just worth it to say what the "generalised binomial theorem" is
Let $q$ be any rational number and $1+a_1x$ be a formal power series. Then we have
\begin{equation}
    (1+a_1x)^q=\sum^{\infty}_{n=0}{q \choose{n}}a_1^nx^n.
\end{equation}
EDIT: Thank you!
 [/INST] Following your steps, we have that
\begin{align}b_n&=[x^n]\frac{5x}{(x-1)^2}=5[x^{n-1}](1-x)^{-2}=5[x^{n-1}]\sum_{k=0}^{\infty}\binom{-2}{k}(-x)^k\\
&=5\binom{-2}{n-1}(-1)^{n-1}=5(-1)^{n-1}\frac{(-2)(-3)\cdot (-2-(n-1)+1)}{(n-1)!}\\&=\frac{5n!}{(n-1)!}=5n.
\end{align}
</s>