I have this recurrence relation:
$$\begin{equation} \begin{cases} a_{n+1} = a_n + n & (n\geq0)\\ a_0 = 0 \end{cases} \end{equation}$$
Set:
$$f(x) = \sum_{n=0}^\infty a_nx^n$$
I have solved:
$$\sum_{n=0}^{\infty}a_{n+1}x^n = \sum_{n=0}^{\infty}a_{n}x^n +\sum_{n=0}^{\infty}nx^n$$ $$\frac{1}{x}\sum_{n=0}^{\infty}a_{n+1}x^{n+1} = f(x) + x \bigg( \frac{1}{(1-x)^2} \bigg)$$ $$\frac{1}{x}f(x) = f(x) + \frac{x}{(1-x)^2} $$ $$f(x) = \frac{x^2}{(1-x)^3}$$
So:
$$f(x) = \frac{x^2}{(1-x)^3} = \frac{A}{1-x}+\frac{B}{(1-x)^2}+\frac{C}{(1-x)^3}$$
And:
$$\begin{equation} \begin{cases} A = 1\\ B = -2\\ C = 1 \end{cases} \end{equation}$$
So I have:
$$f(x) = \frac{x^2}{(1-x)^3} = \frac{1}{1-x}-\frac{2}{(1-x)^2}+\frac{1}{(1-x)^3}$$
How can I procede to get recurrence relation?
You should make use of the fact that $$ \sum_{n=0}^\infty\binom{k+n-1}{n}x^n=\frac{1}{(1-x)^k}; \quad (k\geq 1, |x|<1) $$ which can be obtained by repeatedly differentiating the geometric series or from the generalized binomial theorem. In any case (proceeding from where you left off) $$ f(x)=\sum_{n=0}^\infty \left(1-2(n+1)+\binom{n+2}{2}\right)x^n $$ whence $$ a_n=1-2(n+1)+\binom{n+2}{2};\quad (n\geq 0).$$