It is given that $a_{0}=1$ , $b_{0}=0$ , $c_0=0$
$$ c_n= xc_{n-1}+x(x-1)a_{n-1}(3b_{n-1}+(x-2)a_{n-1}^{2})) $$
$$ b_n=xb_{n-1}+x(x-1)a_{n-1}^{2} $$
$$ a_n=xa_{n-1}+1 $$
where x is any constant.
Is there any method , I can get a formula for $c_n$ in terms of $c_0$ and x?
HINT: $$a_n=\frac{x^{n+1}-1}{x-1}$$ if $x\ne 1$ and inserting this term in the next equation we obtain $$b(n)=x^{n-1} \left(c_1+\frac{x^2 \left(-x^{-n}+x^{n+3}-(2 n+3) x^2+(2 n+3) x\right)}{(x-1)^2}\right)$$