I am trying to solve the following recursive relation $$K_{2i-2}=\frac{a+K_{2i}}{1+aK_{2i}}\quad;\quad K_{2N}=0$$ and $i=1,\cdots,N$
I want to find the solution for $K_{2i}$.
I believe that this relation has to solved from top to bottom. My attempt was to first put $i=N$ that gives $$K_{2N-2}=a$$ then $i=N-1$ gives $$K_{2N-4}=\frac{2a}{1+a^2}$$ then $i=N-2$ gives $$K_{2N-6}=\frac{a+\frac{2a}{1+a^2}}{1+\frac{2a^2}{1+a^2}}=\frac{a^3+3a}{1+3a^2}$$ then $i=N-3$ gives $$K_{2N-8}=\frac{4a^3+4a}{1+6a^2+a^4}$$ then $i=N-4$ gives $$K_{2N-10}=\frac{a^5+10a^3+5a}{1+10a^2+5a^4}$$
I can't find any pattern in it. How can I solve it?
Claim: $$K_{2N-2n} = \frac { \sum_i {n\choose 2i+1} a^{2i+1} } {\sum_i {n\choose 2i} a^{2i}}$$
This is easily proved by induction on $n$.