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 

Let $B$ be the $n\times n$ symmetric matrix defined by $$(B)_{i,i} = 2^{n-2i}+2^{-i},$$ and for $i\neq j$, $$(B)_{i,j} = 2^{n-i-j}.$$ 
Let $A$ be the matrix in your question: $(A)_{i,i}=2^i-1$ and $(A)_{i,j}=-1$ for $i\neq j$. It's not too difficult to check that $BA=I$.
Hence, if $\vec{x}=(x_1,\ldots,x_n)^T$ and $\vec{b}=(2a,\ldots,2^ia,\ldots,2^na)^T$, then your system can be written $$A\vec{x} = \vec{b}$$ and so $\vec{x}= B\vec{b}$, from which you can deduce the formula for $x_i$.
