Let $\alpha$ be a positive integer $\geqslant 5$. Let us consider \begin{align*} &B_1\subseteq \{0,1,2,3,4\}\\ &B_2\subseteq \mathbb{N}-\{0,1,2,3,4\} \end{align*} where $B_2$ can be finite or infinite. How to prove that $$2^\alpha=\sum\limits_{i\in B_1}2^i + \sum\limits_{i\in B_2}2^i$$ is an impossibility ?
Moreover, if not i.e. if the above expression be true, will it be unique expression ?
I was able to show the statement is true for $\alpha=5,6$. But don't know how shall I proceed in general. Can you help me in this regard ?
Edit
Here I am providing my own attempt. Request to all of you to please suggest or correct me if there be any mistake.
First of all, as $\alpha\geqslant 5$ so $B_2$ must not contain any positive integers $\geqslant \alpha$ otherwise the RHS will become at least $2^{\alpha}+1$ making it bigger than LHS.
Now we have $$2^\alpha=\sum\limits_{i\in \mathcal{B}}2^i$$ where $\mathcal{B}\subseteq B_1\cup B_2$.
But even if we consider the ultimate possibility $\mathcal{B}=\{0,1,2,\cdots, \alpha-1\}$ then RHS $=1+2+2^2+\cdots+2^{\alpha-1}=2^\alpha-1<2^\alpha$ which would be contradiction. If $\mathcal{B}$ be a finite subset of $\{0,1,2, \cdots, \alpha-1\}$ we shall still get contardiction.
Please tell me if there be any mistake I have made here
Formally there is possiblity $\mathcal B_1 = \varnothing$, $\mathcal B_2 = \{\alpha\}$. Your argument is correct, but we can prove it directly without appeal to sum of gemoetric sequence.
Let us prove that $2^\alpha = \sum\limits_{i \in \mathcal B} 2^i$ iff $\mathcal B = \{\alpha\}$ using induction on $\alpha$.
If $\alpha = 0$ then it's true, as would there be $\mathcal B \ni n > \alpha$ the sum would be greater then $1$, and would $\mathcal B$ be empty, the sum would be $0$.
If $\alpha > 0$ then $0 \notin \mathcal B$, otherwise the sum would be an odd number. Then let us define $\mathcal C = \{x - 1 | x \in \mathcal B\}$. We have $2^{\alpha - 1} = \frac{2^\alpha}{2} = \frac{\sum\limits_{i \in \mathcal B} 2^i}{2} = \sum\limits_{i \in \mathcal B} 2^{i - 1} = \sum\limits_{i \in \mathcal C}2^i$. By induction it implies $\mathcal C = \{\alpha - 1\}$, therefore $\mathcal B = \{\alpha\}$.