Find value of $a_{2012}$ A sequence $\left\{a_n\right\}$  is defined as:
$a_1=1$, $a_2=2$ and
$$a_{n+1}=\frac{2}{a_n}+a_{n-1}$$ $\forall$ $n \ge 2$
Find $a_{2012}$ 
My Try:
we have
$$a_{n+1}-a_{n-1}=\frac{2}{a_n}$$
$$a_n a_{n+1}-a_{n-1}a_n=2 \tag{1}$$
Replacing $n$ with $n-1$ we get
$$a_{n-1} a_{n}-a_{n-2}a_{n-1}=2 \tag{2}$$ adding $(1)$ and $(2)$
we get
$$a_n a_{n+1}-a_{n-2}a_{n-1}=4 \tag{3}$$
Again replace $n$ with $n-1$ in $(3)$ and adding with $(1)$ we get
$$a_n a_{n+1}-a_{n-2}a_{n-3}=6 \tag{4}$$  Again replace $n$ with $n-1$ in $(4)$ and adding with $(1)$ we get
$$a_n a_{n+1}-a_{n-3}a_{n-4}=8 \tag{5}$$
Continuing the process we get
$$a_na_{n+1}-a_{n-2010}a_{n-2011}=4022$$
Now in above equation put $n=2012$ we get
$$a_{2012}a_{2013}-a_1a_2=4022$$ $\implies$
$$a_{2012}a_{2013}=4024$$
Any further clue?
 A: You already have $$a_n a_{n+1}-a_{n-1}a_n=2$$
So if you set the new sequence $b_k=a_{k+1}a_{k}$ then you get
$$b_k-b_{k-1}=2$$ so,
$b_k$ is an arithmetic sequence with $b_1=a_1a_2=2$ and step $2$. Then
$$b_k=2+2(k-1)=2k$$
then
$$a_{2}a_{1}=2$$
$$4=a_{3}a_{2}$$
$$a_{4}a_{3}=6$$
$$8=a_{5}a_{4}$$
$$...$$
$$2\cdot 2010=a_{2011}a_{2010}$$
$$a_{2012}a_{2011}=2\cdot2011$$
Multiply every equation and get
$$4\cdot8\cdot12\cdot...2020\cdot (a_{2012}\cdot a_1)=2\cdot6\cdot10\cdot...2022$$
$$a_{2012}=\frac{2\cdot6\cdot10\cdot...2022}{4\cdot8\cdot12\cdot...2020}$$
A: By your work we obtain: $$a_{n+2}a_{n+1}=a_{n+1}{a_n}+2,$$
which gives $$a_{n+1}a_n=2+(n-1)2=2n.$$
Thus, $$a_{n}=\frac{2n-2}{\frac{2n-4}{a_{n-2}}}=\frac{n-1}{n-2}a_{n-2},$$
which for even $n$ gives
$$a_n=\frac{(n-1)(n-3)...1}{(n-2)(n-4)...2}a_2=\frac{2(n-1)!!}{(n-2)!!}.$$
Id est, $$a_{2012}=\frac{2\cdot2011!!}{2010!!}.$$
A: If you put $b_n=a_na_{n+1}$, your relation (1) give that $b_n-b_{n-1}=2$; from this you easily prove that $b_n=2n$ for all $n$, ie $a_na_{n+1}=2n$. Now  $a_{n+2}=2(n+1)/a_{n+1}=(1+1/n)a_n$, and if $c_k=a_{2k}$, you have $c_{k+1}=\frac{2k+1}{2k} c_k$; it is easy to finish. 
