Find the value of the the term

Question

The sequence $a_1,a_2,a_3,\ldots$ satisfies $a_1=1$, $a_2=2$, and $$a_{n+2}=\frac2{a_{n+1}}+a_n\;;$$ find the value of

$$\frac{a_{2012}2^{2009}}{2011}$$

Answer 1

HINT:

$$a_{n+2}=\frac2{a_{n+1}}+a_n\iff a_{n+2}a_{n+1}=a_{n+1}a_n+2$$

$$\implies a_{n+2}a_{n+1}=2n+a_2a_1=2n+2\ \ \ \ (1)$$

$$\text{Replacing }n\text{ with } n-1,\implies a_{n+1}a_n=2n$$

$$\implies a_{n+2}a_{n+1}\cdot2n=a_{n+1}a_n(2n+2)$$

$$\implies a_{n+2} =a_n\frac{(n+1)}n$$

$$n=2m-2\implies a_{2m} =a_{2m-2}\frac{(2m-1)}{2m-2}$$

$$ a_{2m} =\frac{(2m-1)(2m-3)\cdots3}{(2m-2)(2m-4)\cdots4}a_2$$

$$\text{Now,} \frac{(2m-1)(2m-3)\cdots3}{(2m-2)(2m-4)\cdots4}=2\frac{(2m-1)(2m-2)\cdots4\cdot3\cdot2\cdot1}{2^{m-1}\cdot (m-1)! ((2m-2)\cdot4\cdot2)}=\frac{(2m-1)!}{2^{2m-2} \{(m-1)!\}^2}$$ and $a_2=2$

Answer 2

As lab bhattacharjee shows, $a_{n+2}a_{n+1}=2n+2$, and thus, $$ \begin{align} \frac{a_{n+2}}{a_n} &=\frac{a_{n+2}}{a_{n+1}}\frac{a_{n+1}}{a_n}\\ &=\frac{n+1}{n} \end{align} $$ Therefore, $$ \begin{align} \frac{a_{2n}}{a_2} &=\frac{2n-1}{2n-2}\frac{2n-3}{2n-4}\cdots\frac32\\ &=\color{#C00000}{\frac1{4^{n-1}}\frac{(2n-1)!}{(n-1)!(n-1)!}}\\ &=\frac{n}{2^{2n-1}}\binom{2n}{n} \end{align} $$ and since $a_2=2$, $$ a_{2n}=\frac{n}{4^{n-1}}\binom{2n}{n} $$ I believe this verifies lab's hint (in red).