<s>[INST] 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?
 [/INST] 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
$a_{n+1} = \frac{3^{n+2} - 3}{3^{n+2}-1}$
</s>