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10:26 PM
Suppose
$$
u_{n+1}=\frac1{\lambda-u_n}\tag1
$$
after a few iterations, we see that we get something looking like
$$
u_n=\frac{r_n-s_nu}{p_n-q_nu}\tag2
$$
and that given a form as in $(2)$,
$$
\begin{align}
u_{n+1}
&=\frac1{\lambda-\frac{r_n-s_nu}{p_n-q_nu}}\tag{3a}\\
&=\frac{p_n-q_nu}{\lambda(p_n-q_nu)-(r_n-s_nu)}\tag{3b}\\
&=\frac{p_n-q_nu}{(\lambda p_n-r_n)-(\lambda q_n-s_n)u}\tag{3c}
\end{align}
$$
Therefore,
$$
r_{n+1}=p_n\quad\text{and}\quad s_{n+1}=q_n\tag{4a}
$$
and
$$
p_{n+1}=\lambda p_n-r_n\quad\text{and}\quad q_{n+1}=\lambda q_n-s_n\tag{4b}
Here is a lengthy, but not complicated, derivation of a general solution.
 
11:06 PM
In your question, set $b_n=\sqrt3\,a_n$ and then set $\lambda=\frac4{\sqrt3}$
then $p_n=\frac{\sqrt3}2\left(\left(\sqrt3\right)^n-\left(\frac1{\sqrt3}\right)^n\right)$ and $q_n=\frac{\sqrt3}2\left(\left(\sqrt3\right)^{n-1}-\left(\frac1{\sqrt3}\right)^{n-1}\right)$
$a_n=\frac1{\sqrt3}\frac{\left(\left(\sqrt3\right)^{n-1}-\left(\frac1{\sqrt3}\right)^{n-1}\right)-\left(\left(\sqrt3\right)^{n-2}-\left(\frac1{\sqrt3}\right)^{n-2}\right)\sqrt3\,a_1}{\left(\left(\sqrt3\right)^n-\left(\frac1{\sqrt3}\right)^n\right)-\left(\left(\sqrt3\right)^{n-1}-\left(\frac1{\sqrt3}\right)^{n-1}\right)\sqrt3\,a_1}$
 
11:33 PM
$a_n=\frac{\left(3^{n-1}-1\right)-\left(3^{n-1}-3\right)a_1}{\left(3^n-1\right)-\left(3^n-3\right)a_1}$
This is the answer that you quoted here.
 

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