Solve a linear system of equation involving some recursion $$
\begin{align*}
x_{1} &= 1 + x_{2}\\
x_{2} &= 1 + \frac{1}{2} x_{3} + \frac{1}{2} x_{1}\\
&\vdots\\
x_{i} &= 1 + \frac{1}{2} x_{i+1} + \frac{1}{2} x_{1}\\
&\vdots\\
x_{n-2} &= 1 + \frac{1}{2} x_{n-1} + \frac{1}{2} x_{1}\\
x_{n-1} &= 1 + \frac{1}{2} x_{n} + \frac{1}{2} x_1 \\
x_{n} &= 0 \\
\end{align*}
$$
EDIT: I found one way to solve this. It's simply plugging successive $x_i$ into the first equation. When you reach $x_n$, since $x_n = 0$, you end up with an equation of just $x_1$ which you can solve (using geometric sum) to get $x_1 = 3\times 2^{n-2} - 2$. The rest then is easy. 
If you have a more elegant solution, please share.

Disclaimer
I'm gonna write couple of terms and then general equation for both forward and backward substitution, so you'll need to use mathematical induction to actually prove them
First, substitute $x_1$ to the second equation
$$
2 x_2 = 2 + x_3 + x_1 = 2 + x_3 + 1 + x_2 = 3 + x_3 + x_2 \implies x_2 = 3\cdot 2^0 + x_3
$$
Now, substitute both $x_2$ and $x_1$ to the equation for $x_3$
$$
2x_3 = 2 + x_4 + x_1 = 2 + x_4 + 1 + x_2 = 3 + x_4 + 3 + x_3 = 3 \cdot 2^1 + x_4 + x_3 \implies x_3 = 3 \cdot 2^1 + x_4
$$
Now you can prove (using mathematical induction) that
$$
x_{n-1} = 3 \cdot 2^{n-3} + x_n
$$
Now, do backward substitution by using $x_n = 0$ 
$$
x_{n-1} = 3 \cdot 2^{n-3}
$$
and then
$$
x_{n-2} = 3 \cdot 2^{n-4} + x_{n-1} = 3 \cdot 2^{n-4} + 3 \cdot 2^{n-3} = 3 \cdot 2^{n-4} (2^0 + 2^1)
$$
and one more
$$
x_{n-3} = 3 \cdot 2^{n-5} + x_{n-2} = 3 \cdot 2^{n-5} + 3 \cdot 2^{n-4}(2^0 + 2^1) = 3 \cdot 2^{n-5} (2^0 + 2^1 + 2^2)
$$
Now, you can prove that
$$
x_{n - k} = 3 \cdot 2^{n-k-2} (2^0 + 2^1 + \ldots + 2^{k-1})
$$
In the parenthesis is nothing but simple geometric progression, so
$$
x_{n-k} = 3 \cdot 2^{n-k-2} (2^k - 1)
$$
You can do that all the way until $k = n - 2$ to find $x_2$
$$
x_2 = 3 \cdot 2^0 (2^{n-2} - 1) = 3(2^{n-2}-1)
$$
and finally
$$
x_1 = 1 + x_2 = 3 \cdot 2^{n-2} - 2
$$
Summary
\begin{align}
x_1 &= 3 \cdot 2^{n-1} - 2 \\
x_i &= 3 \cdot 2^{i-2} (2^{n-i} - 1)\quad \text{for}\ 2 \le i < n\\
x_n &= 0
\end{align}
