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riddles >> putnam exam (pure math) >> Proof by Mathematical Induction
(Message started by: daemonturk on Sep 12th, 2009, 7:51am)

Title: Proof by Mathematical Induction
Post by daemonturk on Sep 12th, 2009, 7:51am
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.

Title: Re: Proof by Mathematical Induction
Post by towr on Sep 12th, 2009, 10:42am
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

Title: Re: Proof by Mathematical Induction
Post by french_math on Jun 9th, 2010, 3:17am
This is quite easy :

1+2+...+n-1 = (n-1)*n/2, that you can prove by induction too.



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