TITLE: Is it true that $\sum_{i=0}^{n-1} 2^i$ divides $\sum_{i=n}^{2n-1} 2^i$?
QUESTION [6 upvotes]: Some investigation suggested to me that
$$\sum_{i=0}^{n-1} 2^i \Bigg| \sum_{i=n}^{2n-1} 2^i$$
The point being that binary numbers in the form exclusively of an even number of $1$s appear to be divisible by the number represented by just half that number of $1$s.
Is this the case, and if so why is that please?

REPLY [2 votes]: An alternative answer, by construction:
In binary, multiplying a number by 2 is equivalent to shifting all of its bits one place higher and adding a zero on the end. Consequently, multiplying by $2^n$ just means doing that $n$ times, moving the bits $n$ places higher and adding $n$ zeros.
The object is to prove that a number consisting of $n$ consecutive ones and $n$ consecutive zeros is divisible by the number consisting of $n$ consecutive ones.
Knowing what we know about multiplication by $2$ in binary, it should be straightforward how to get from the smaller number to the larger one: just shift its digits $n$ places higher and add $n$ zeros; in other words, multiply by $2^n$. Since we can multiply the smaller number by an integer to get to the larger number, this means that the larger number is divisible by the smaller one.
In fact, this method can be used to prove a stronger statement: a number consisting of $n$ consecutive ones followed by any number of consecutive zeros is divisible by a number consisting of $n$ consecutive ones.