Finding the coefficient of $x^{50}$ in $\frac{(x-3)}{(x^2-3x+2)}$ First, the given answer is: $$-2 + (\frac{1}{2})^{51}$$
I have tried solving the problem as such:
$$[x^{50}]\frac{(x-3)}{(x^2-3x+2)} = [x^{50}]\frac{2}{x-1} + [x^{50}]\frac{-1}{x-2}$$ 
$$ = 2[x^{50}](x-1)^{-1} - [x^{50}](x-2)^{-1}$$ 
$$=2\binom{-1}{50}-\binom{-1}{50} = \binom{-1}{50} = \binom{50}{50} = 1$$
which is different from the correct answer. Can anyone tell me what I'm doing wrong here?
Edit:
As Did mentions in the comments, $$2[x^{50}](x-1)^{-1} = -2\binom{-1}{50} = -2 \neq 2\binom{-1}{50}$$
Also, $$- [x^{50}](x-2)^{-1} = [x^{50}]\frac{1}{2-x} = \frac{1}{2}[x^{50}]\frac{1}{1-\frac{x}{2}}$$
$$ = \frac{1}{2}[x^{50}](1-\frac{x}{2})^{-1} = \frac{1}{2}\binom{-1}{50}(\frac{-1}{2})^{50}$$
$$=(\frac{1}{2})^{51}$$
Both added together gives the correct answer, which is the correct solution using the original method.

The beginning looks good, but I do not see how you justify the last line. I would use the geometric series instead:
$$\begin{align*}\frac{x-3}{x^2-3x+2} &= \frac{2}{x-1} - \frac{1}{x-2}\\
&= -2\frac{1}{1-x} + \frac{1}{2}\frac{1}{1-\frac 12 x} \\
& = -2 \sum_{n=0}^\infty x^n + \frac{1}{2}\sum_{n=0}^\infty \frac{x^n}{2^n}
\end{align*}$$
From that we can easily see that $$[x^{50}]\frac{x-3}{x^2-3x+2} = -2+\frac{1}{2^{51}}$$
