Find the sequence. I am given generating function $\ f(x)=\frac{1}{x-3} $
I have to find formula of the sequence which is generated by this function.
Have to achieve this?
I find some tutorials but it was hard for me to understand this.
In some of them people where starting from generating function for $a_{n}=1$ (I mean $ \sum_{n=0}^\infty x^n $ ) and then made transformation to achieve desired result. I would be extremely grateful if someone would explain me how to do this.
 A: Note that 
$$\frac{1}{x-3}=-\frac{1}{3}\cdot\frac{1}{1-x/3}.$$
You probably know that (for suitable $t$)
$$\frac{1}{1-t}=1+t+t^2+t^3+\cdots.$$
Plug in $t=x/3$ and simplify a bit to get the expansion of your function. 
A: You're looking for a power series
$$y = \sum_{n = 0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$$
such that $y = \frac{1}{x - 3}$.  Or equivalently such that $(x - 3)y = 1$.  Well we have
$$xy = x\sum_{n = 0}^\infty a_nx^n = \sum_{n = 0}^\infty a_nx^{n + 1} = \sum_{n = 1}^\infty a_{n - 1}x^n$$
and
$$3y = 3\sum_{n = 0}^\infty a_nx^n = \sum_{n = 0}^\infty 3a_nx^n$$
so
$$xy - 3y = \sum_{n = 1}^\infty a_{n - 1}x^n - \sum_{n = 0}^\infty 3a_nx^n = \sum_{n = 1}^\infty a_{n - 1}x^n - \sum_{n = 1}^\infty 3a_nx^n - 3a_0$$
$$= \sum_{n = 1}^\infty[a_{n - 1} - 3a_n]x^n - 3a_0$$
For this to equal $1$ we must have that the coefficient of $x^0$ which is $3a_0$ equal $1$, so $a_0 = \frac13$.  For $n > 0$ the coefficient of $x^n$, which is $a_{n - 1} - 3a_n$ must equal $0$.  So $a_n = \frac{a_{n - 1}}{3}$.  Hence $a_1 = \frac{1}{3^2}$, $a_2 = \frac{1}{3^3}$, and so on.  Can you figure out a formula for $a_n$?
A: To achieve this, you expand the function in a power series around the origin. The coefficients of this power series form the sequence you seek.
