To prove by induction that all other lights are on given that light k=1 is on, we need to show that if light k=n is on, then lights 2n and 2n+1 are also on for any positive integer n. Here's the proof:
Base Case: When n = 1, we have light k=1 on. We need to show that lights 2(1)=2 and 2(1)+1=3 are also on.
Indeed, if light k=1 is on, then lights 2(1)=2 and 2(1)+1=3 are also on, as per the given condition.
Inductive Step: Assume that for some positive integer k = n, if light k=n is on, then lights 2n and 2n+1 are also on.
Now, we need to prove that if light k = n+1 is on, then lights 2(n+1) and 2(n+1)+1 are also on.
Since light k = n+1 is on, by the induction hypothesis, we know that lights 2n and 2n+1 are on.
Now, we can show that lights 2(n+1) and 2(n+1)+1 are on:
Light 2(n+1) = 2n+2 is on. This is true because 2n is on by the induction hypothesis, and the given condition states that if light k is on, then 2k is also on.
Light 2(n+1)+1 = 2n+3 is on. This is true because 2n+1 is on by the induction hypothesis, and the given condition states that if light k is on, then 2k+1 is also on.
Thus, if light k = n+1 is on, then lights 2(n+1) and 2(n+1)+1 are also on.
By the principle of mathematical induction, we have shown that if light k=1 is on, then all other lights (lights 2, 3, 4, 5, ...) are also on.