You could just start by proving that any natural number can be written as (2^k)n+2^(k-1)-1, and then prove that if a number can be written like that, then it has that property. That should make it obvious that every natural number has that property, without thinking about infinite unions.
The problem that I have though isn't in the form of 2
kn+2
k-1-1. I just used that as an example. The problem I have is of the form 2
kn + c, where c takes on a whole bunch of values that don't seem to have any rhyme or reason. (I've found a way to compute c, but it's really really complicated). However, the number of c's that work for a specific k seems to be easier to find, so I was looking for another way to do it. My idea was to use the number of c's for a given k (let's call it c
k) divided by 2
k (this fraction is basically the fraction of natural numbers it includes) and summing them up:
(c
k1)/(2
k1) + (c
k2)/(2
k2) + (c
k3)/(2
k3) + (c
k4)/(2
k4) + ...
Also, when I say a fraction of the natural numbers, say a/b, I'm saying that if you picked b consecutive natural numbers, there would be a of them.
So how would I prove that the union of all of these subsets is the natural numbers?