$a_{n+2}=\frac{2}{a_{n+1}}+a_{n}.$ and $a_{1}=1,a_{2}=2$.Then value of $a_{2017}$ 
If ${a_{n}}$ be a sequence defined by $\displaystyle a_{n+2}=\frac{2}{a_{n+1}}+a_{n}.$ and $a_{1}=1,a_{2}=2$.Then value of $a_{2017}$

Try: $$a_{n+2}a_{n+1}-a_{n}a_{n+1}=2\frac{a_{n+1}}{a_{n-1}}$$
$$\sum^{2015}_{n=1}\left(a_{n+2}a_{n+1}-a_{n}a_{n+1}\right)=2\sum^{2015}_{n=1}\left(\frac{a_{n+1}}{a_{n-1}}\right)$$
$$a_{2017}a_{2016}-a_{2}a_{3}=a_{2015}a_{2016}$$
Could some help me to solve it, Thanks

Hint: apply a telescoping sum to
$$
a_{n+2}a_{n+1}-a_{n+1}a_n=2
$$

The solution of the equation above is $a_{n+1}a_n=2n$
$$
\begin{align}
\frac{a_{2n+2}}{a_{2n}}
&=1+\frac{2}{a_{2n+1}a_{2n}}\\
&=1+\frac1{2n}
\end{align}
$$
Therefore,
$$
\begin{align}
a_{2n+2}
&=2\prod_{k=1}^n\frac{2n+1}{2n}\\
&=\frac{4n+2}{4^n}\binom{2n}{n}
\end{align}
$$
Since $a_{2n+2}a_{2n+1}=4n+2$, we get
$$
a_{2n+1}=\frac{4^n}{\binom{2n}{n}}
$$
Thus,
$$
a_{2017}=\frac{4^{1008}}{\binom{2016}{1008}}
$$
