calculus - Find formula for $a_{2n}$ and $a_{2n-1}$ of recursive sequence $a_n$

Take for example the sequence $a_n=1+\frac{1}{a_{n-1}}$ with $a_1=1$ which is defined recursively.

How do I find the formula for $a_{2n}$ and $a_{2n+1}$? Is there a general approach to find the formulas for any recursively defined sequence $a_n$?

I found a formula saying $a_{2n} = 2 - \frac{1}{1 + a_{n-1}}$ which seems to work out. And by definition of $a_n$, $a_{2n+1} = 1 + \frac{1}{a_{(2n+1)-1}} = 1 + \frac{1}{a_{2n}}$ and the first formula can be used. But I don't understand how $a_{2n}$ was found in the first place.

Answer

..for any recursively defined sequence $a_n$? No, you are clearly asking too much, especially because there are chaotic sequences, like the logistic map. On the other hand, every sequence behaving like

$$a_{n+1} = \frac{Aa_n+B}{Ca_n+D}$$
has a simple closed form just depending on the powers of the $\begin{pmatrix}A& B \\ C & D\end{pmatrix}$ matrix.
Have a look at this similar question. In your case, you may notice that the matrix
$$\begin{pmatrix}1 & 1 \\ 1 & 0\end{pmatrix}$$
has eigenvalues $\varphi,\overline{\varphi}$ and check by induction that $a_n=\frac{F_{n+1}}{F_{n}}$ is associated with the convergents of the continued fraction $\varphi=[1;1,1,1,1,\ldots]$.