Thread: Need help with the algebra in an induction proof

Hi, I've been working on this problem for quite a while now and I just can't seem to get the algebra working for it. Any help is appreciated.

Determine the exact set of natural numbers n for which the inequality 2^n>=(n+1)^2 holds (1).

I have already found that this is true for n>=6 from intuition and plugging stuff in, but I need to prove this by induction.

Base case: n=6 holds for both sides

Induction step: I need to prove that (1) holds for n=k+1.
I tried plugging k+1 in for the left side, yielding 2^(k+1) = 2^k*2^1>=2*(k+1)^2 (from (1)) But this gets me nowhere as my ultimate goal is to arrive at 2^(k+1)>=((k+1)+1)^2
Any ideas?

Hello morbius27

Originally Posted by morbius27

Hi, I've been working on this problem for quite a while now and I just can't seem to get the algebra working for it. Any help is appreciated.

Determine the exact set of natural numbers n for which the inequality 2^n>=(n+1)^2 holds (1).

I have already found that this is true for n>=6 from intuition and plugging stuff in, but I need to prove this by induction.

Base case: n=6 holds for both sides

Induction step: I need to prove that (1) holds for n=k+1.
I tried plugging k+1 in for the left side, yielding 2^(k+1) = 2^k*2^1>=2*(k+1)^2 (from (1)) But this gets me nowhere as my ultimate goal is to arrive at 2^(k+1)>=((k+1)+1)^2
Any ideas?

Suppose that is the propositional function:

Then

, since when

So

Grandad