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

[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.

 

[Gunthi]

2^(k+1)=2^k*2>=2*(k+1)^2 (by induction hypothesis).

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

 

[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 don't 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.

 

[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.