$a_{n+2}=\frac{2}{a_{n+1}}+a_{n}.$ and $a_{1}=1,a_{2}=2$.Then value of $a_{2017}$

Question

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

Answer 1

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

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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, $$ a_{2017}=\frac{4^{1008}}{\binom{2016}{1008}} $$