Expanding a generating function in a series

Question

For a given recurrence relation the generating function is A(x)=$\frac{x}{(1-x)(1-2x)}$.

Then the book says that if we want to find an explicit formula for the $a_n$'s we would have to expand A(x) in a series. The partial fraction of A(x) is $$ x \{\frac{2}{1-2x}-\frac{1}{1-x}\}$$ and this is clear to me. But how did the author expand this $$ x \{\frac{2}{1-2x}-\frac{1}{1-x}\} = \{2x+2^2x^2+2^3x^3+...\}-\{x+x^2+x^3+..\} $$

Hope my question is clear. Thanks in advance for your help.

Answer 1

Hint: $$\frac{x}{(1-x)(1-2x)}=\frac{1}{1-2x}-\frac{1}{1-x}$$

Answer 2

Hint: $\frac{1}{1-x} =1+x+x^2 +... $ for $x<1$.

Answer 3

It only took me a few more minutes to get it. From the geometric series evaluation, we know that $\sum_nx^n=\frac{1}{1-x}$ for $|x|=1$.

And know taking the summation of our partial fraction : $ \frac{1}{1-2x}-\frac{1}{1-x}$ we get

$$ \sum_n \frac{1}{1-2x}-\frac{1}{1-x}=\sum_n\frac{1}{1-2x}-\sum_n\frac{1}{1-x} = (2x)^n - x^n$$ and hence we have $$\{2x+2^2x^2+2^3x^3+...\}-\{x+x^2+x^3+...\}$$