, 2012

We have
$$x_2 = \frac{2}{4} \left( \frac{1}{6} + \frac{1}{2} \right) = \frac{2}{6}, \quad x_3 = \frac{3}{5} \left( \frac{2}{6} + \frac{1}{2} \right) = \frac{3}{6}, \text{ etc.}$$
We will prove by induction that for every $n \ge 1$, we have $x_n = \frac{n}{6}$.
Indeed, assuming $x_n = \frac{n}{6}$, it follows
$$x_{n+1} = \frac{n+1}{n+3} \left( \frac{n}{6} + \frac{1}{2} \right) = \frac{n+1}{n+3} \cdot \frac{n+3}{6} = \frac{n+1}{6},$$
and we are done.
The answer is $x_{2011} = \frac{2011}{6}$.