Finding $a_n$ when $a_1=5$, $a_n a_{n+1}=2a_n-1$ $(n \in \mathbb{N})$ (Question Edited)

Question

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 $a_1=5$ , $a_n a_{n+1}=2a_n−1$

Answer 1

$$\begin{align} \because & a_na_{n+1}=2a_n-1\\ \therefore & a_n(a_{n+1}-1)=a_n-1\\ \therefore & \frac{a_n}{a_n-1}=\frac{1}{a_{n+1}-1}\\ \therefore & \frac 1{a_n-1}+1=\frac 1{a_{n+1}-1} \end{align}$$ Ok,let $b_n=\frac 1{a_n-1}$, it is so easy.Now

$$b_1=\frac 1{a_1-1}=\frac 14\quad b_n+1=b_{n+1}\quad\Longrightarrow\quad \frac 1{a_n-1}=b_n=n-\frac 34 \quad\Longrightarrow\quad a_n=1+\frac 1{n-\frac 34}=\frac {4n+1}{4n-3}$$

Answer 2

It is clear that $a_1=5=\dfrac{4\times1+1}{4\times1-3}$.

Now, if $a_n=\dfrac{4n+1}{4n-3}$, then\begin{align}a_{n+1}&=\frac{2a_n-1}{a_n}\\&=2-\frac1{a_n}\\&=2-\frac{4n-3}{4n+1}\\&=\frac{4n+5}{4n+1}\\&=\frac{4(n+1)+1}{4(n+1)-3}.\end{align}

Answer 3

Hint: Let $b_n=\frac{4n+1}{4n-3}$. Prove that $b_1=5$, $b_n b_{n+1}=2b_n-1$. Conclude that $b_n = a_n$.

Bottom line: a recurrence of order $1$ is determined by the recurrence law and the initial condition.