How find this linear equation $-x_{1}-x_{2}-x_{3}-\cdots+(2^n-1)x_{n}=2^na$?

Question

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

Answer 1

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 $1\leqslant k \leqslant n$.

Answer 2

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 $x_i$.