Given a vector $(n\_0, n\_1, \dots, n\_l)$ where $n\_i \in \{-1, 1\}$, $i = \overline{0, l-1}, n\_l = 1$ and $l \in \mathbb{N}$.  

Prove that for all $a$ such that
$$0 < a \leq 2^0\cdot n\_0 + 2^1 \cdot n\_1 + \dots + 2^{l - 1} \cdot n\_{l - 1} + 2^l \cdot n\_l$$


there are distinct $k\_0, k\_1, \dots, k\_r \in I = \{0, 1, \dots, l\}$ where $r \leq l$ such that 


$$a = n\_{k\_0} \cdot 2^{k\_0} + n\_{k\_1} \cdot 2^{k\_1} + \dots + n\_{k\_r} \cdot 2^{k\_r}\;.$$ 


Any help is appreciated.



Powers of two with coefficients {1,−1}
Any $N\ge0$ has a unique binary expansion 
$$
N=\sum\_{k=0}^n a\_k 2^k, \quad a\_k\in\{0,1\},\ a\_n=1.
$$
(Here $n=\lfloor \log\_2 N\rfloor$.)
Now consider the map $N\mapsto 2N-(2^{n+1}-1)$ from $\mathbb Z\_+$ to $\mathbb Z$. We have
$$
N'=2N-(2^{n+1}-1)=\sum\_{k=0}^n a'\_k 2^k, \quad a'\_k\in\{-1,1\},
$$
where $a\_k'=2a\_k-1$. This map is 1-1, hence the result. (So, in fact, we can expand negative integers in such a series as well as positive ones.)


