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?

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, $$a_{2012}=\frac{2\cdot2011!!}{2010!!}.$$
