TITLE: Finding $a_n$ when $a_1=5$, $a_n a_{n+1}=2a_n-1$ $(n \in \mathbb{N})$ (Question Edited)
QUESTION [5 upvotes]: When $a_1=5$, $a_n a_{n+1}=2a_n-1$ $(n \in \mathbb N)$, 
I inducted that $a_n=\frac{4n+1}{4n-3}$ it by manually calculating $a_1, a_2, a_3...$ and finding out the pattern of the numerator and denominator. But it doesn't seem that accurate- how can I deduct $a_n$ from the given conditions?
Edit: I'm really sorry- I didn't write my question clearly. $a_n=\frac{4n+1}{4n-3}$ isn't one of the 'given conditions'. The question is deducting '$a_n=\frac{4n+1}{4n-3}$' itself from $a_1=5$ , $a_n a_{n+1}=2a_n−1$

REPLY [2 votes]: $$\begin{align}
\because  & a_na_{n+1}=2a_n-1\\
\therefore & a_n(a_{n+1}-1)=a_n-1\\
\therefore & \frac{a_n}{a_n-1}=\frac{1}{a_{n+1}-1}\\
\therefore & \frac 1{a_n-1}+1=\frac 1{a_{n+1}-1}
\end{align}$$
Ok,let $b_n=\frac 1{a_n-1}$, it is so easy.Now
$$b_1=\frac 1{a_1-1}=\frac 14\quad b_n+1=b_{n+1}\quad\Longrightarrow\quad \frac 1{a_n-1}=b_n=n-\frac 34 \quad\Longrightarrow\quad a_n=1+\frac 1{n-\frac 34}=\frac {4n+1}{4n-3}$$