Saudi Arabia Mathematical Competitions 2012

We have $$x_2=\frac{2}{4}\left(\frac{1}{6}+\frac{1}{2}\right)=\frac{2}{6},x_3=\frac{3}{5}\left(\frac{2}{6}+\frac{1}{2}\right)=\frac{3}{6},\text{ etc.}$$ 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 $x_{2011}=\frac{2011}{6}$.