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Topic: Proof by Mathematical Induction (Read 20241 times) 

daemonturk
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Proof by Mathematical Induction
« on: Sep 12^{th}, 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
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Re: Proof by Mathematical Induction
« Reply #1 on: Sep 12^{th}, 2009, 10:42am » 
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The base case is simple 1^{2} = 1^{3}, so it is true for n=1. Now assume it is true for n1, so (1+2+3+...+n1)^{2}=1^{3}+2^{3}+3^{3}+...+(n1)^{3}, then to prove it holds for n, you have to prove that you can go from this to (1+2+3+...+n)^{2}=1^{3}+2^{3}+3^{3}+...+n^{3}. If you expand the latter a little, you have (1+2+3+...+n1)^{2} + 2 n(1+2+3+..n1) + n^{2} = 1^{3}+2^{3}+3^{3}+...+(n1)^{3} + n^{3} Therefore, to account for the change from the case of n1 to n, we need to prove that n^{2} + 2 * n*(1+2+3+..n1) = n^{3}

« Last Edit: Sep 12^{th}, 2009, 10:43am by towr » 
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french_math
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Re: Proof by Mathematical Induction
« Reply #2 on: Jun 9^{th}, 2010, 3:17am » 
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This is quite easy : 1+2+...+n1 = (n1)*n/2, that you can prove by induction too.


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