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Talk:Squared triangular number

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It would be nice to include the picture proof from http://users.tru.eastlink.ca/~brsears/math/oldprob.htm#s32, since it is clearer than the one already on this page, but I am not sure about the copyright status of the images. Please advise.

Loeb 02:17, 30 October 2007 (UTC)[reply]

It is a pretty proof. I think only the second of the two images there can easily stand on its own. BUT, I would like to know the source of the proof. Has it been published anywhere? Is it the same as one of the existing proofs already known in the article? I don't think a web page makes an adequate reference for a problem with as much history as this one. —David Eppstein 02:22, 30 October 2007 (UTC)[reply]

Proof

Who was the first to prove this theorem? I've seen it called Nicomachus' Theorem, but did he prove it or merely observe it? If not, did Āryabhaṭa? Al-Karajī?

CRGreathouse (t | c) 18:22, 31 August 2009 (UTC)[reply]

Sum of the first "n" cubes - even cubes - odd cubes (geometrical proofs)

The "IDEA"
New method for summing the first "n" cubes
Sum of the first "n" even cubes
Sum of the first "n" odd cubes


Starting from the basic idea described in the first animation, we introduced a new procedure to obtain formulas for summing the first "n" cubes, even cubes and odd cubes. This method, called "Successive Transformations Method", consists in an inductive handling of a geometric model, in order to obtain another equivalent which gives evidence of the searching formulas.




See the animations.





Consider the final transformation that you see in the second animation, and we compare this figure with the square base parallelepiped that contains it. We expect that the ratio between these figures becomes 1/2 to infinity. Performing calculations with Excel, you see that this is true. In addition, as has happened in the discussion at the link /media/wikipedia/commons/6/6c/Pubblicazione_english.pdf, you encounter these other amazing results: 

      n
 lim (Σn n3)/(Σn n).n2 = 1/2     
 n→∞   1
      n
 lim (Σn n5)/(Σn n).n4 = 1/3     
 n→∞   1
      n
 lim (Σn n7)/(Σn n).n6 = 1/4     
 n→∞   1
      n
 lim (Σn n9)/(Σn n).n8 = 1/5     
 n→∞   1

which, by induction, can be generalized in a formula. Note that the denominators of the results are the positions of the exponents in the numerator in the sequence of odd numbers. The induction principle enshrines the validity of this "theorem". --79.17.53.156 (talk) 09:21, 3 August 2013 (UTC)[reply]

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