In this applet, we use 3D blocks to represent an important identity of Number Theory.

[color=#0000ff][i]note[/i][/color]: you can directly open this material in [i]GeoGebra 3D Graphing[/i] just clicking in [url=https://www.geogebra.org/3d/pe9wgzjs]https://www.geogebra.org/3d/pe9wgzjs[/url]

[color=#0000ff][i]note[/i][/color]: you can directly open this material in [i]GeoGebra 3D Graphing[/i] just clicking in [url=https://www.geogebra.org/3d/u332rret]https://www.geogebra.org/3d/u332rret[/url]

It is well known that [math]1+2+3+\ldots+n=\frac{n\cdot\left(n+1\right)}{2}.[/math] Also well known, but more difficult to prove that [math]1\cdot1+2\cdot2+3\cdot3+\ldots+n\cdot n=\frac{n\cdot\left(n+1\right)\cdot\left(2n+1\right)}{6}.[/math] Usually this latter formula is proven by using induction, but it hides the geometrical background of the right side of the equation.[br]Recently I read a "proof without words" [url=http://www.maa.org/sites/default/files/Siu15722.pdf]explanation[/url] of this important formula, published by Man-Keung Siu from the University of Hong Kong. [br]The applet is limited to [math]n\le3[/math] for technical reasons. Although, if you download the material, in the desktop version you may want to increase [math]n[/math]. Nevertheless, the proof is still easy to understand even if [math]n=3[/math]. For all cases, you have to consider the height of each level is 1.

Actually, the proven formula is [math]1\cdot1+2\cdot2+3\cdot3+\ldots+n\cdot n=\frac{n\cdot(n+\frac{1}{2})\cdot(n+1)}{3}[/math], which is equivalent to the previous one.