If $4 = \frac{3}{a_1} = \frac{3}{a_2}+a_1= \frac{3}{a_3} + a_2 = ... = \frac{3}{a_{n+1}} + a_n$ Prove that $a_n= \frac{3^{n+1} - 3}{3^{n+1}-1}$ If $$4 = \frac{3}{a_1} = \frac{3}{a_2}+a_1= \frac{3}{a_3} + a_2 = \space ... \space = \frac{3}{a_{n+1}} + a_n$$
Prove that $$a_n= \frac{3^{n+1} - 3}{3^{n+1}-1}$$
I tried to prove this via induction, but i't didn't work. Any hints with this problems?
 A: You want to prove
$$a_n= \frac{3^{n+1} - 3}{3^{n+1}-1} \tag{1}\label{eq1}$$
From $4 = \frac{3}{a_1}$, you have $a_1 = \frac{3}{4}$, which matches \eqref{eq1} when $n = 1$ as it gives $a_1 = \frac{9 - 3}{9 - 1} = \frac{6}{8} = \frac{3}{4}$. Assume \eqref{eq1} holds for all $n \le k$ for some integer $k \ge 1$. Now, from the final part of your definition,
$$4 = \frac{3}{a_{n+1}} + a_n \; \implies \; 4 - a_n = \frac{3}{a_{n+1}} \; \implies \; a_{n+1} = \frac{3}{4 - a_n} \tag{2}\label{eq2}$$
Using $n = k$, along with the induction hypothesis, gives
\begin{align}
a_{k+1} & = \frac{3}{4 - a_k} \\
& = \frac{3}{4 - \frac{3^{k+1} - 3}{3^{k+1} - 1}} \\
& = \frac{3\left(3^{k+1} - 1\right)}{4\left(3^{k+1} - 1\right) - \left(3^{k+1} - 3\right)} \\
& = \frac{3^{k+2} - 3}{(4-1)3^{k+1} - 4 + 3} \\
& = \frac{3^{k+2} - 3}{3^{k+2} - 1} \tag{3}\label{eq3}
\end{align}
Note that this matches the right side of \eqref{eq1} for $n = k + 1$, so it's definition of $a_{k+1}$ matches. Thus, this completes the induction step, proving \eqref{eq1} is true for all $n \ge 1$.
A: Proof by induction
Base case
$a_1 = \frac 43$
Suppose our proposition is true.
$\frac{3}{a_{n+1}} + a_n = 4\\
a_{n+1} = \frac{3}{4-a_n}\\
a_{n+1} = \frac{3}{4-\frac{3^{n+1} - 3}{3^{n+1} - 1}}$
By the inductive hypothesis
$a_{n+1} = \frac{3^{n+2} - 3}{3^{n+2}-1}$
