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The rest came by itself.--[[User:Ancora Luciano|Ancora Luciano]] ([[User talk:Ancora Luciano|talk]]) 18:47, 14 May 2013 (UTC)
:That does indeed seem to be a valid and nice proof (or at least something that could be made into a proof with a little more care about why the limit of the saw-tooth areas converges to the parabola area (in contrast to situations like [http://mathworld.wolfram.com/DiagonalParadox.html this one] for which the convergence argument doesn't work). But either it's not new or it doesn't (yet) belong here; see [[WP:NOR]]. So if you think it's new, the appropriate thing to do would be to get it properly published elsewhere first, so that the publication can be used as a [[WP:RS|reliable source]] here. Personally maintained web sites are not adequate for this purpose. —[[User:David Eppstein|David Eppstein]] ([[User talk:David Eppstein|talk]]) 22:48, 14 May 2013 (UTC)
== Geometrical derivation of the formula ==
[[File:Pyram.jpg|thumb|tridimensional model]]
We represent the pyramid number P <sub> 6 </ sub> = 91 with cubes of unit volume, as shown, and inscribe in building a pyramid (in red). Let V <sub> 6 </ sub> the volume of the inscribed pyramid. To obtain P <sub> 6 </ sub> you may add to V <sub> 6 </ sub> the excess external volume to the red pyramid. Such excess is: 2/3 for each cube placed on the central edge, and 1/2 for the cubes forming the steps of the building (enlarge for a better look of highlighted part). Then, calculating one has:
:P<sub>6</sub> = V<sub>6</sub>+(2/3)*6+(1+2+3+4+5)
: For the induction principle, will be:
:P<sub>n</sub> = V<sub>n</sub>+(2n)/3+Σ(from 1 to n-1)n
:P<sub>n</sub> = n<sup>3</sup>/3+2n/3+(n<sup>2</sup>+n)/2-n
:P<sub>n</sub> = (2n<sup>3</sup>+3n<sup>2</sup>+n)/6
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