How find this linear equation $-x_{1}-x_{2}-x_{3}-\cdots+(2^n-1)x_{n}=2^na$? let $a\in R$,then solve this follow equation:

$$\begin{cases}
x_{1}-x_{2}-x_{3}-\cdots-x_{n}=2a\\
-x_{1}+3x_{2}-x_{3}-\cdots-x_{n}=4a\\
-x_{1}-x_{2}+7x_{3}-\cdots-x_{n}=8a\\
\cdots\cdots\cdots\cdots\cdots\cdots\\
-x_{1}-x_{2}-x_{3}-\cdots+(2^n-1)x_{n}=2^na
\end{cases}$$

Maybe this problem have more methods,Thank you 

Write $s = x_1 + x_2 + \dotsc + x_n$. Then the system becomes
$$2^k x_k - s = 2^k a;\quad 1 \leqslant k \leqslant n.$$
Rearranging that, we obtain
$$x_k - a = 2^{-k}s.$$
Summing these equations yields
$$\sum_{k=1}^n (x_k-a) = s - na = \left(\sum_{k=1}^n 2^{-k}\right)s = \left(1 - 2^{-n}\right)s.$$
From that we obtain
$$s = n2^na$$
and
$$x_k = a(1 + n2^{n-k})$$
for $1\leqslant k \leqslant n$.
