Recursive sequence (German Math Olympiad 2016, 2nd round)

Question

Let sequence $x_1, x_2, x_3, ...$ be recursively defined by

$x_{n+1} = \dfrac{n+1}{n+3} \left(x_n + \dfrac{1}{2}\right)$

and let $x_1 = \dfrac{1}{6}$

Calculate $x_{2016}$

How would you approach this? Find a closed form?

Answer 1

Hint: $$x_2 = \frac{1}{3} $$ and $$x_{n+1}= \dfrac{n+1}{n+3} \left(x_n + \dfrac{1}{2}\right) = \dfrac{n+1}{n+3} \left(\dfrac{n}{n+2} \left(x_{n-1} +\frac{1}{2} \right)+ \dfrac{1}{2}\right)$$ $$\Longrightarrow (n+3)(n+2)x_{n+1} = (n+1)nx_{n-1} +(n+1)^2$$ Denote $y_n=(n+2)(n+1)x_{n}$, then $$y_2 = 4.3.x_2 = 4 \tag{1}$$ $$y_{2n}=y_{2(n-1)} + 4n^2 \tag{2}$$

From $(1),(2)$, you can deduce easily $y_{2016}$ and $x_{2016}$ is just equal to $\frac{y_{2016}}{2017.2018}$.