Sum of recursive series I am taking a course in algebra and this problem was on my problem set, and I had no idea how to solve it.
Suppose we have a sequence $s_n$ of real numbers such that $5s_{n+1}-s_{n}-3s_{n}s_{n+1}=1$ for $1 \leq n \leq 42$ and $s_1=s_{43}$. 
What are the possible values of $s_1+s_2+ \ldots + s_{42}$?
 A: Note that we have 
$$s_{n+1} (5-3s_n) = 1 + s_n,$$
thus in particular $s_n \neq 5/3$ for all $n$ and so 
$$\tag{1} s_{n+1} = \frac{1+s_n}{5-3s_n}.$$
The mapping 
$$f(z) = \frac{z + 1}{-3z+5}$$
is a Mobius transform. One can check that 
$$f^{(n)} (z) = \frac{az+ b}{cz+ d},$$
where 
$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ -3 & 5\end{bmatrix}^n$$
Doing some linear algebra, we have 
$$\begin{bmatrix} 1 & 1 \\ -3 & 5\end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 3\end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & 4\end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 3\end{bmatrix}^{-1}$$
When $n=42$, we have (direct computations)
$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 3\cdot 2^{42} -4^{42} & -2^{42} + 4^{42} \\ 3\cdot 2^{42} - 3\cdot 4^{42} & 3\cdot 4^{42} - 2^{42} \end{bmatrix}$$
So the fact that $s_1 = s_{43}$ is the same as 
$$s_1 = \frac{(3\cdot 2^{42}-4^{42}) s_1 +(-2^{42} + 4^{42})}{(3\cdot 2^{42} - 3\cdot 4^{42})s_1 + 3\cdot 4^{42} - 2^{42}}$$
By some calculations, 
$$3s_1^2 - 4s_1 + 1 = 0\Rightarrow s_1= 1 \text{ or } \frac 13.$$
Put into $(1)$, we get either $s_n = 1$ for all $n$ or $s_n = \frac 13$ for all $n$. So 
$$s_1 + \cdots + s_{42} = 42 \text{ or } 14.$$
A: Let $u_n = {3s_n-1 \over 1-s_n}$ (unless $s_n=1$ which gives us sum $42$). Then $u_{n+1} = {3{1+s_n \over 5-3s_n}-1 \over 1-{1+s_n \over 5-3s_n}} = {2 \over 4}{3s_n-1 \over 1-s_n} = {1 \over 2}u_n$; $u_{43}=u_1$ implies $u_1=0, s_1={1 \over 3}$.
