This is the definition of the series ($a,b,x \in \mathbf{R}$):
$$s_{n+1} = \frac{s_n+ax}{1-bx} $$
I would like to find a closed formula, to calculate $s_n$ from $s_0$
How is it possible? The denominator makes it impossible to factor out $s_n$, and I have no idea how to progress further.
Written less fancy, you have $$ s_{n+1}=\alpha s_n+\beta$$ with $\alpha=\frac{1}{1-bx}$ and $\beta=\frac{ax}{1-bx}$. If we let $t_n=s_n-\gamma$, we find $$t_{n+1}-\gamma= \alpha t_n+\beta-\alpha\gamma $$ so that if we are lucky and $\gamma=\alpha\gamma-\beta$ (i.e., $\gamma=\frac{\beta}{\alpha-1}=\frac{a}{b}$, at least when no division by $0$ occurs), we have $$t_{n+1}= \alpha t_n$$ and clearly $$ t_n=\alpha^nt_0.$$ This gives us $$ s_n=\alpha^nt_0+\gamma = \alpha^n(s_0-\gamma)+\gamma =\frac{s_0-\frac ab}{(1-bx)^n}+\frac ab.$$ I leave ist to you to check what happens when $bx=1$ or $bx=0$.
By induction easily we can show that $$s_n = \frac{s_0 + ax + ax\sum_{i=0}^{n-2}(1-bx)^{2^i}}{(1-bx)^{2^{n-1}}}$$ If we simplify the above equation we can get the above equation too.
