<s>[INST] 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.
 [/INST] 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}
</s>