Alternative problem statement:
Prove that any natural number can be written as $$\sum_{i = 0}^{n} a_{i} 2^i$$ if $ a_{i}\in[0, 1]$ for some $n \in \mathbb{N}$.
Further information:
$a_{i}$ is NOT a constant. $a_{i}$ can take any integer value within the interval $[0, 1]$ (zero or one) at choice for every $i$ (with $0 \leq i \leq n$ and both $i, n \in \mathbb{N}$). We include the zero inside $\mathbb{N}$ when mentioning the natural numbers' set.
Generalization of previous problem:
Prove that any natural number can be written as $$\sum_{i = 0}^{n} a_{i} b^i$$ if $a_{i} \in [0, b-1]$ at choice per every $i$ and if $b-2, n, i \in \mathbb{N}$.
For 0, we just need $a_0 = 0$. For 1, $a_0 = 1$ suffices.
Let's call the positive number to be represented $m_0$.
Choose $n$ such that $2^n \leq m_0 < 2^{n + 1}$. Now compute $m_1 = m_0 - 2^n$ and assign $a_n = 1$. What do we know about $m_1$? Only that it is less than $n$. It can have any value from 0 to $2^n - 1$. If $m_1 > 0$, then we can find some power of 2 just like we did for $m_0$, but the exponent will be smaller than $n$. We can just keep repeating this process until eventually winding up with 0. And if we keep track of the $a_i$, we have the binary representation of $m_0$, which is what almost all computers today use to represent non-negative integers.
I'm not sure if you correctly notated the interval. Since $2 \times 2^\alpha = 2^{\alpha + 1}$, assigning $a_i = 2$ would be inefficient at best, an algorithmic error at worst.