Q:

Find the n-th number from the generating function

Is there any way to find the n-th number in the series, by knowing it's genereting function.
For example, I found that the closed form solution for a generating function $\displaystyle\frac{(1+x)^2}{(1-x)^3}$ can be expressed as $2n^2 + 2n + 1$.
I am interested how can I derive similar formula
$\displaystyle\frac{(1+x)^5}{(1-x)^6}$.

A:

In other words, you want to find the coefficient of $x^n$ in the Maclaurin series for the function
$$f(x)=\frac{(1+x)^5}{(1-x)^6}=(1+x)^5(1-x)^{-6}.$$
To do that, first expand each factor using the binomial formula
$$(1+u)^m=\sum_{n=1}^\infty\binom mnu^n$$
and then multiply the two binomial series together using
$$\left(\sum_{n=0}^\infty a_nx^n\right)\cdot\left(\sum_{n=0}^\infty b_nx^n\right)=\sum_{n=0}^\infty c_nx^n\text{ where }c_n=a_0b_n+a_1b_{n-1}+\cdots+a_nb_0.$$ 
For your function
$$f(x)=(1+x)^5(1-x)^{-6}$$
I get
$$\sum_{k=0}^5\binom5k\binom{n+5-k}5$$
as the coefficent of $x^n$. For the function $(1+x)^2(1-x)^{-3}$ the corresponding formula is
$$\sum_{k=0}^2\binom2k\binom{n+2-k}2=\binom{n+2}2+2\binom{n+1}2+\binom n2=2n^2+2n+1.$$

