Suppose $a_{1}=a_{2}=\frac{1}{3}$, for $n = 3, 4, 5, ...$ , we have $$a_{n}=\frac{a_{n-1}^2(1−2a_{n-2})}{2a_{n-1}^2 − 4a_{n-2}a_{n-1}^2+a_{n-2}}$$
(1) Prove that $\frac{1}{a_{n}} −2$ are square numbers
(2) Find out the general term formula of ${a_{n}}$
I have done the first part using induction by assuming $x_n=\frac{1}{a_n} -2$, we get $$x_n=\frac{1}{x_{n-2}a_{n-1}^2}$$ on simplifying the relation but not able to solve to get general term, Can anyone suggest, how to proceed?
It seems that $\big(\frac{1}{a_n}-2\big)^{1/2}$ is sequence A001835 . Can you prove this by induction?
Find everything you need to know there. Many references. And in particular $$ \big(\frac{1}{a_n}-2\big)^{1/2} = \frac{(3 + \sqrt3)^{2n - 1} + (3 - \sqrt3)^{2n - 1}}{6^n} $$