Geometric interpretation for sum of fourth powers

Summing the first n first powers of natural numbers:
nk=1k=12n(n+1)
and there is a geometric proof involving two copies of a 2D representation of (1+2++n) that form a n×(n+1)-rectangle.


Similarly,
nk=1k2=16n(n+1)(2n+1)
has a geometric proof (scroll down just a bit til you see the wooden blocks) involving six copies of a 3D representation of (12+22++n2) that form a n×(n+1)×(2n+1)-rectangular solid.


And similarly,
nk=1k3=14n2(n+1)2
has a proof involving four copies of a 4D representation of (13+23++n3) that form a n×n×(n+1)×(n+1)-rectangular hypersolid in four-space. I can’t find a resource to demonstrate this last one better, but I’ve sketched it out for n=3, sketching the four dimensions using a 4×4 grid for two dimensions, and within each cell a 3×3 grid for the other two. (Try it – it’s fun!) Here’s a smaller n=2 version. Note that 13+23 is represented by some 4D blocks 1×13+1×23 that are configured in a way that makes use of all four dimensions.

This is a 2×2×3×3 rectangular hypersolid made up of four copies of 113+123.


So we move on to fourth powers:
nk=1k4=130n(n+1)(2n+1)(3n2+3n1)
This time the polynomial does not completely factor. I would like to know if anyone can find a similar geometric proof. It would involve 30 copies of a 5D representation of (14+24++n4) forming a 5D hypersolid, one of whose 2-faces somehow works out to have area 3n2+3n1, with the orthogonal edge lengths being n, n+1, and 2n+1. I know it seems hopeless…30 copies? 3n2+3n1? But it would make for some nice art.

It would be a nice start if there were some sort of connection between 30 and a symmetry group of R5. For instance, if there were a subgroup of SO5 whose order was some large divisor of 30, then that might help place the 30 blocks of volume 15 into their configuration.

Answer

On this book:
http://www.amazon.com/Proofs-without-Words-Exercises-Classroom/dp/0883857006

Is given the following solution:

Proofs Without Words

I guess that with a bit of calculation it is possible to get to the formula for ni=1i4

I know that we miss all the stuff about the fifth dimension… but in this way the proof is much easier to visualize…

Attribution
Source : Link , Question Author : alex.jordan , Answer Author : ThePunisher

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