- #1
morbius27
- 14
- 0
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.
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.