2^n > n^2 for n> 4 where n is a natural number

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Question from John, a student:

I've been asked to prove this:
2^n > n^2 for n> 4 and n is a natural number

But I can't get past the part. So
(i) let n = 5, then 2^5 > 5^2
32 > 25 , true.

(ii) assume statement is true for n, prove it also holds for n+1

then I have 2^(n+1) > ( n+1)^2

I put it in the form 2^n * 2 > n^2 + 2n + 1

but I don't know how to use induction to make use of my assumption to prove this statement.

Any help would be appreciated.

John,

Suppose we know k² < 2^k for some k ≥ 5, then (k+1)² = k²(1+1/k)² < 2^k(1+1/5)² =2^k(36/25) < 2^k+1.

Penny

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