# Proving 2^n>=(n+1)^2 by mathematical induction

1. Feb 23, 2010

### morbius27

Hello, In this problem I am trying to Determine the exact set of natural numbers n for which the inequality 2^n>=(n+1)^2 holds. (equation (1))

I have already dealt with the base case where n=6, (since the inequality does not hold for n<6), and so (1) holds for n=k, and I need to show that it holds for n=k+1. It's the algebra that I really can't figure out. I tried 2^(k+1)=2^k*2>=2*(k+1)^2 (by induction hypothesis).

I need to get this in the form 2^(k+1)>=((k+1)+1)^2, but I have no idea how to get to this point from here. Any help would be much appreciated.

2. Feb 23, 2010

### Gunthi

Your induction hypothesis is wrong, you have to consider n=k+1 in both sides of the inequality that you have as the thesis, $$2^n\geq(n+1)^2$$, wich is what you are looking for.

3. Feb 24, 2010

### morbius27

well I know you have to start with one side of the inequality by plugging in k+1 for n, and then manipulating it to the point where you can apply the induction hypothesis (2^n>=(n+1)^2) and then more manipulation to get to 2^(k+1)>=((k+1)+1)^2...I just dont know which side of the inequality to start with (2^(k+1) or ((k+1)+1)^2 and the necessary algebra to arrive at the conclusion.

4. Feb 24, 2010

### Gunthi

I don't usually do as you say, but for me it is much easier to put k+1 in both therms and try one of two things, to get to a true expression where the initial inequation is evident, or to try to get to the induction hypothesis from the initial equation also.

In both of these ways I can always be sure to be getting the right thing because the initial inequation, that we assume as true, is always there.

Try this way, if you get stuck in the algebra just ask.