At time 0, a particle resides at the point 0 on the real line. Within 1 second, it divides into 2 particles that fly in opposite directions and stop at distance 1 from the original particle. Within the next second, each of these particles again divides into 2 particles flying in opposite directions and stopping at distance 1 from the point of division, and so on. Whenever particles meet they annihilate (leaving nothing behind).
How many particles will there be at time $2^1$$^1$ + 1?
Here I observed that at time $2^n$ there would be two particles and at time time $2^n$ + 1. There would be 4 particles. But don't know how to prove it. I was trying with mathematical Induction but couldn't do it. Please help.
Using induction, prove that at $n=2^k$, there are two particles at positions $x=-2^k$ and $x=2^k$.
Base case: at $n=2^1$, there are two particles at positions $x=-2$ and $x=2$.
Inductive step: if there are two particles at positions $x=-2^k$ and $x=2^k$ when $n=2^k$, then:
After another $2^k$ seconds those two particles will have each split multiple times, but in systems that would not have interacted until the last split, because for one system every particle had $x<0$, but for the other every particle had $x>0$. At the $(2^k)^{th}$ second, each system would have resulted in two particles (each), but two of the particles collide at $x=0$ and annihilate. Therefore, there are two particles at time $n=2^k + 2^k = 2^{k+1}$, located at $x=-2^{k+1}$ and $x=2^{k+1}$.