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Fiesta28 [93]
3 years ago
11

Prove 2^n > n for all n equal to or greater than 1. I mostly need help with how to solve the problem when it is greater than

rather than equal to. Thanks for the help
Mathematics
1 answer:
noname [10]3 years ago
7 0
If n is an integer, you can use induction. First show the inequality holds for n=1. You have 2^1=2>1, which is true.

Now assume this holds in general for n=k, i.e. that 2^k>k. We want to prove the statement then must hold for n=k+1.

Because 2^k>k, you have

2^{k+1}=2\times2^k>2k

and this must be greater than k+1 for the statement to be true, so we require

2k>k+1

for k>1. Well this is obviously true, because solving the inequality gives 3k>1\implies k>\dfrac13. So you're done.

If you n is any real number, you can use derivatives to show that 2^n increases monotonically and faster than n.
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