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Topic: Proof by Mathematical Induction (Read 20245 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
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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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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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