<s>[INST] How to solve recurrence relations using generating functions that do not start at $a_0$? For example,
$$
a_{n+1} - 2a_n = 2, \\ \textrm{for } n \geq 1, a_1 = 1
$$
I tried using
$$
\sum_{n=1}^\infty a_{n+1}x^{n+1} - \sum_{n=1}^\infty 2a_n x^{n+1} = \sum_{n=1}^\infty 2x^{n+1}
$$
if we let $f(x) = \sum_{n=1}^\infty a_n x^n$
$$
f(x) - a_1 x - 2x f(x) = \frac{2x^2}{1-x}
$$
and I get 
$$
f(x) = \frac{x^2 + x}{(1-x)(1-2x)} = \frac{A}{1-x} + \frac{B}{1-2x}
$$
but the partial fraction decomposition is not giving me a solution. Also, using method of undetermined coefficients I got the right answer which is
$$
a_n = 2^{n-1}3 - 2
$$
edit1: typos
 [/INST] Should the fraction on the RHS be 
$$ \frac{2x^2}{1-x}\text{?}$$
edit: You should end up with 
$$f(x) = \frac{2x^2}{(1-2x)(1-x)} + \frac{x}{(1-2x)}$$
which you can simplify and do partial fractions with and get the desired answer.
</s>