``` wu :: forums (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi) 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>=1Need a speedy response. Title: Re: Proof by Mathematical Induction Post by towr on Sep 12th, 2009, 10:42am The base case is simple12 = 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 + n3Therefore, to account for the change from the case of n-1 to n, we need to prove thatn2 + 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. Powered by YaBB 1 Gold - SP 1.4! Forum software copyright © 2000-2004 Yet another Bulletin Board