TITLE: What is the general term of $a_{n+1}=\frac{2a_n-1}{5a_n-1} \ , \ \ a_1=1$?
QUESTION [6 upvotes]: I've struggled to solve this exercise

$$a_{n+1}=\frac{2a_n-1}{5a_n-1}\ , \ \ a_1=1$$
$$b_{n+1}=(5a_n-1)b_n \ , \ \ b_1=1$$
Find $b_{\ 40}$ .

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I thought 'taking inverse' will be useful, but not yet... :-(
$\color{red}{01.}$ How can I find $b_{\ 40}$ ?
$\color{red}{02.}$ How can I find the general term of $a_n$, $b_n$ ?
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Thank you for your attention to this matter.
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REPLY [5 votes]: since
$$b_{n+1}=(5a_{n}-1)b_{n}\Longrightarrow a_{n}=\dfrac{1}{5}\left(\dfrac{b_{n+1}}{b_{n}}+1\right)$$
take in
$$a_{n+1}=\dfrac{2a_{n}-1}{5a_{n}-1}\Longrightarrow \dfrac{1}{5}\left(\dfrac{b_{n+2}}{b_{n+1}}+1\right)=\dfrac{\frac{2}{5}\left(\dfrac{b_{n+1}}{b_{n}}+1\right)-1}{\frac{b_{n+1}}{b_{n}}}$$
$$\Longrightarrow \dfrac{b_{n+2}+b_{n+1}}{5b_{n+1}}=\dfrac{\frac{2}{5}b_{n+1}+\frac{2}{5}b_{n}-b_{n}}{b_{n+1}}$$
$$\Longrightarrow b_{n+2}+b_{n+1}=2b_{n+1}-3b_{n}$$
$$\Longrightarrow b_{n+2}=b_{n+1}-3b_{n},b_{1}=1,b_{2}=(5a_{1}-1)b_{1}=4$$
The corresponding characteristic equation is
$$r^2=r-3\Longrightarrow r=\dfrac{1\pm\sqrt{11}i}{2}$$
so
$$b_{n}=A\left(\dfrac{1+\sqrt{11}i}{2}\right)^n+B\left(\dfrac{1-\sqrt{11}i}{2}\right)^n$$
since $b_{1}=1,b_{2}=4$
so we have
$$\begin{cases}
A\left(\dfrac{1+\sqrt{11}i}{2}\right)+B\left(\dfrac{1-\sqrt{11}i}{2}\right)=1\\
A\left(\dfrac{1+\sqrt{11}i}{2}\right)^2+B\left(\dfrac{1-\sqrt{11}i}{2}\right)^2=4
\end{cases}$$
$$\Longrightarrow \begin{cases}
3A+B\left(\dfrac{1-\sqrt{11}i}{2}\right)^2=\dfrac{1-\sqrt{11}i}{2}\\
A\left(\dfrac{1+\sqrt{11}i}{2}\right)^2+3B=\dfrac{1+\sqrt{11}i}{2}
\end{cases}\Longrightarrow \begin{cases}
A=\dfrac{7+\sqrt{11}i}{\sqrt{11}i-11}\\
B=\dfrac{\sqrt{11}i-7}{11+\sqrt{11}i}
\end{cases}$$
so
$$b_{n}=\dfrac{7+\sqrt{11}i}{\sqrt{11}i-11}\cdot\left(\dfrac{1+\sqrt{11}i}{2}\right)^n+\dfrac{\sqrt{11}i-7}{11+\sqrt{11}i}\cdot\left(\dfrac{1-\sqrt{11}i}{2}\right)^n$$