Since the equation is homogeneous: We can solve for $x_{n+1}$ in terms of $x_n$ to get $x_{n+1}=\frac{a\pm\sqrt{a^2+4b}}{2}x_n$ So in general, for each $n$ there is a $k$ such that $x_n=\left(\frac{a+\sqrt{a^2+4b}}{2}\right)^k\left(\frac{a-\sqrt{a^2+4b}}{2}\right)^{n-k}x_0=s^kt^{n-k}x_0$ So there are usually $n+1$ possible values for $x_n$. An exception is that if $a=0$ makes there be only $2$ possible values for $x_n$. Finally, $S_n=\sum_{k=0}^ns^kt^{n-k}x_0=\frac{s^{n+1}-t^{n+1}}{s-t}x_0=\frac{\left(a+\sqrt{a^2+4b}\right)^{n+1}-\left(a-\sqrt{a^2+4b}\right)^{n+1}}{2^{n+2}\sqrt{a^2+4b}}x_0$ This sum obeys the recursion: $\begin{cases}S_0=x_0\\S_1=a\\S_n=aS_{n-1}+bS_{n-2}\end{cases}$