Recurrence and Fibonacci: $a_{n+1}=\frac {1+a_n}{2+a_n}$ For the recurrence relation 
$$a_{n+1}=\frac {1+a_n}{2+a_n}$$
where $a_1=1$, the solution is $a_n=\frac {F_{2n}}{F_{2n+1}}$, where $F_n$ is the $n$-th Fibonacci number, according to the convention where $F_1=0, F_2=1,\cdots
$
This can easily be proven by substituting the solution back into the recurrence relation, or alternatively by induction. 

Can the solution be derived directly  from the recurrence relation itself, i.e.  not by substituting the solution into the recurrence, or by induction?


[My interpretation of Mike Earnest's solution, posted here for my own reference only]
$$\begin{align}
a_{n+1}&=\frac {b_{n+1}}{c_{n+1}}=\frac {1+\frac {b_n}{c_n}}{2+\frac {b_n}{c_n}}=\frac {b_n+c_n}{b_n+2c_n}\\\\
\left(b_{n+1}\atop c_{n+1}\right)
&=\left(1\ \ 1\atop 1\ \  2\right)\left(b_{n}\atop c_{n}\right)
=\left(1\ \ 1\atop 1\ \  2\right)^2\left(b_{n-1}\atop c_{n-1}\right)=\cdots\\\\
&=\left(1\ \ 1\atop 1\ \  2\right)^n\left(b_1\atop c_1\right)\\\\
&=\left(0\ \ 1\atop 1\ \  1\right)^{2n}\left(b_1\atop c_1\right)\\\\
&=\left(F_{2n-1}\ \ F_{2n}\ \ \ \atop F_{2n}\ \ \ \  F_{2n+1}\right)\left(1\atop 1\right)
&&\scriptsize(a_1=1\Rightarrow b_1=1, c_1=1)\\\\
&=\left(F_{2n-1}+F_{2n}\ \ \ \atop F_{2n}\ \ \ +F_{2n+1}\right)\\\\
&=\left(F_{2n+1}\atop F_{2n+2}\right)\\\\
\therefore a_{n+1}&=\frac {b_{n+1}}{c_{n+1}}=\frac {F_{2n+1}}{F_{2n+2}}\\\\
\color{red}{a_n}&\color{red}{=\frac {F_{2n}\ \ \ }{F_{2n+1}}\qquad\blacksquare}
\end{align}$$
