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Topic: Proof by Mathematical Induction (Read 20241 times) |
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daemonturk
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Proof by Mathematical Induction
« on: Sep 12th, 2009, 7:51am » |
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Use proof by mathematical induction to prove that: (1+2+3+...+n)^2=1^3+2^3+3^3+...+n^3 for n>=1 Need a speedy response.
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towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
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Re: Proof by Mathematical Induction
« Reply #1 on: Sep 12th, 2009, 10:42am » |
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The base case is simple 12 = 13, so it is true for n=1. Now assume it is true for n-1, so (1+2+3+...+n-1)2=13+23+33+...+(n-1)3, then to prove it holds for n, you have to prove that you can go from this to (1+2+3+...+n)2=13+23+33+...+n3. If you expand the latter a little, you have (1+2+3+...+n-1)2 + 2 n(1+2+3+..n-1) + n2 = 13+23+33+...+(n-1)3 + n3 Therefore, to account for the change from the case of n-1 to n, we need to prove that n2 + 2 * n*(1+2+3+..n-1) = n3
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« Last Edit: Sep 12th, 2009, 10:43am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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french_math
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Re: Proof by Mathematical Induction
« Reply #2 on: Jun 9th, 2010, 3:17am » |
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This is quite easy : 1+2+...+n-1 = (n-1)*n/2, that you can prove by induction too.
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