# Proof by InductionWatch

#1
Prove for every positive integer

Now I've done with the basis case, but the assume n = k part. Say I can rearrange to get something like this, , problem is if u compare this with and expand it, I cannot prove that . Any ideas? I thought of taking a method of difference approach i.e but I couldn't come up with anything. Help!
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10 years ago
#2
You're supposed to prove that if then .
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10 years ago
#3
Hold on - you're trying to get (2k + 1)^2 - 1, not what you wrote. This is 4k^2 + 4k.
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10 years ago
#4
Hmm, I'm not seeing a straightforward way to do this. It can be shown that , but in order to complete the proof from there it would be necessary to show that for all positive integers k... There's probably an easier way I'm not thinking of.
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10 years ago
#5
One way:

From 3^n > 2n^2 - 1 we get

3^(n+1) > 3(2n^2-1). Then consider 3(2n^2-1)-(2(n+1)^2-1) and show it's > 0 for n > 1.
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10 years ago
#6
Ah, of course. I did that but when the completed-square form came out as I thought I was going the wrong way, since I thought the condition required was > 0 for .
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10 years ago
#7
In these cases, never forget you don't have to start from n = 1. If you end up only being able to show for n > 3, there's nothing stopping you checking explicitly for n = 1, 2, 3, 4.

It's not uncommon to have to do this sort of thing for real problems: Bertrand's postulate says that there's always a prime between n and 2n. One proof shows analytically that there's always a prime between n and 2n for n >= 512. Then the proof for n < 512 is done by direct calculation.
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