# Geometric interpretation for sum of fourth powers

Summing the first $n$ first powers of natural numbers: $$\sum_{k=1}^nk=\frac12n(n+1)$$ and there is a geometric proof involving two copies of a 2D representation of $(1+2+\cdots+n)$ that form a $n\times(n+1)$-rectangle.

Similarly, $$\sum_{k=1}^nk^2=\frac16n(n+1)(2n+1)$$ has a geometric proof (scroll down just a bit til you see the wooden blocks) involving six copies of a 3D representation of $(1^2+2^2+\cdots+n^2)$ that form a $n\times(n+1)\times(2n+1)$-rectangular solid.

And similarly, $$\sum_{k=1}^nk^3=\frac14n^2(n+1)^2$$ has a proof involving four copies of a 4D representation of $(1^3+2^3+\cdots+n^3)$ that form a $n\times n\times(n+1)\times(n+1)$-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 $4\times4$ grid for two dimensions, and within each cell a $3\times 3$ grid for the other two. (Try it - it's fun!) Here's a smaller $n=2$ version. Note that $1^3+2^3$ is represented by some 4D blocks $1\times1^3+1\times2^3$ that are configured in a way that makes use of all four dimensions. $$\begin{array}{c|c|c} \begin{array}{cc} {\color{red} \bullet} & {\color{red} \bullet}\\ {\color{red} \bullet} & {\color{red} \bullet} \end{array} & \begin{array}{cc} {\color{red} \bullet} & {\color{red} \bullet}\\ {\color{red} \bullet} & {\color{red} \bullet} \end{array} & \begin{array}{cc} {\color{blue} \bullet} & {\color{blue} \bullet}\\ {\color{blue} \bullet} & {\color{blue} \bullet} \end{array} \\\hline \begin{array}{cc} {\color{green} \bullet} & {\color{green} \bullet}\\ {\color{green} \bullet} & {\color{green} \bullet} \end{array} & \begin{array}{cc} {\color{red} \bullet} & {\color{blue} \bullet}\\ {\color{green} \bullet} & {\color{magenta} \bullet} \end{array} & \begin{array}{cc} {\color{blue} \bullet} & {\color{blue} \bullet}\\ {\color{blue} \bullet} & {\color{blue} \bullet} \end{array} \\\hline \begin{array}{cc} {\color{green} \bullet} & {\color{green} \bullet}\\ {\color{green} \bullet} & {\color{green} \bullet} \end{array} & \begin{array}{cc} {\color{magenta} \bullet} & {\color{magenta} \bullet}\\ {\color{magenta} \bullet} & {\color{magenta} \bullet} \end{array} & \begin{array}{cc} {\color{magenta} \bullet} & {\color{magenta} \bullet}\\ {\color{magenta} \bullet} & {\color{magenta} \bullet} \end{array} \\ \end{array}$$ This is a $2\times2\times3\times3$ rectangular hypersolid made up of four copies of $1\cdot1^3+1\cdot2^3$.

So we move on to fourth powers: $$\sum_{k=1}^nk^4=\frac1{30}n(n+1)(2n+1)(3n^2+3n-1)$$ 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 $(1^4+2^4+\cdots+n^4)$ forming a 5D hypersolid, one of whose 2-faces somehow works out to have area $3n^2+3n-1$, with the orthogonal edge lengths being $n$, $n+1$, and $2n+1$. I know it seems hopeless...30 copies? $3n^2+3n-1$? 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 $\mathbb{R}^5$. For instance, if there were a subgroup of $\operatorname{SO}_5$ whose order was some large divisor of $30$, then that might help place the $30$ blocks of volume $1^5$ into their configuration.

• +1. But I want no part in this madness... Nov 21, 2012 at 10:46
• In the same way that there are nice 'visual' proofs of $\sum_kk^3$ that instead go through its representation as $\left(\sum_kk\right)^2$, maybe you can wedge 5 copies of $\sum_kk^4$ together and find $(3n^2+3n-1)$ copies of $\sum_kk^2$ in the result? Nov 21, 2012 at 17:55
• Are you familiar with the book: Charming Proofs: A Journey into Elegant Mathematics (Dolciani Mathematical Expositions)? I believe there are others also for visualizing proofs and they are very interesting indeed. Nov 21, 2012 at 18:28
• 30 comes from the fact that the Bernoulli number $B_4=-1/30$. See "Faulhaber's formula" in Wikipedia. Apr 23, 2013 at 0:26
• @pharmine, Yes and so does $6$ in the sum of squares formula. But perhaps coincidentally there is a way to configure 6 shapes together as seen in the link. I'm asking if coincidentally this can be done with thirty 5-dimensional shapes. (I doubt that it can, but I'd love to see it if I'm wrong.) Apr 23, 2013 at 4:16

Is given the following solution:

I guess that with a bit of calculation it is possible to get to the formula for $$\sum_{i=1}^ni^4$$

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

• +1 This is cool, but I'm holding out for five-dimensional blocks :) Oct 21, 2013 at 20:26

Your construction sounds like that of the Ehrhart polynomials. Snooping around Wikipedia a bit gives us confirmation. From the article square pyramidal number:

In modern mathematics, figurate numbers are formalized by the Ehrhart polynomials.

The Ehrhart polynomial $L(P,t)$ of a polyhedron $P$ is a polynomial that counts the number of integer points in a copy of $P$ that is expanded by multiplying all its coordinates by the number $t$. The Ehrhart polynomial of a pyramid whose base is the unit square $[0,1]^2 \times 0$, and whose apex is an integer point at height one above the base plane $(0,0,1)$ has Ehrhart polynomial $\frac{1}{6}(t+1)(t+2)(2t+3)$.

In your problem the choice of convex polytope is pretty clear. It is the convex hull of the hypercube $[0,1]^4 \times 0$ and the point $(0,0,0,0,1)$.

Reading the article Ehrhart Polynomial we get some definitions:

Let $P$ be a polytope with vertices in a lattice $L$ (e.g. $L = \mathbb{Z}^d$ then) $L(P,t) = \# (tP \cap L)$ is a polynomial in $t$, called the Erhart polynomial of degree $d = \dim L$.

The Ehrhart polynomial of the interior (those strictly inside $P$) is $L(\text{int}(P), t)= (-1)^d L(P,-t)$.

• But a convex polytope is not an analogy to blocks. With blocks in the smaller dimensional cases, we can see directly the connection to $1^k+2^k+\cdots+n^k$. The polytope answer to this question for summing $1+2+\cdots+n$ would be two triangles glued together instead of the familiar interlocking stairsteps. Mar 12, 2015 at 20:33
• Well, yes, $1^k+2^k+...+n^k$ is the value of an Ehrhart polynomial for the 'hybercubic pyramid'. But does this help to compute the sum? Mar 12, 2015 at 21:06
• @alex.jordan Did you notice the sum of squares divides sum of 4th powers? $$\sum k^2 =\frac{1}{6}n(n+1)(2n+1) \Bigg| \sum k^4 = \frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)$$ The blog you mentioned has two solutions. One uses a triangular lattice. $$.$$ The volume of a pyramid is $\frac{1}{3}Ah$ where $A$ is the base and $h$ is the height. By Cavalieri principle we don't worry if the Pyramid is slanted or not. Mar 12, 2015 at 21:13
• I think you are confused about what the question asks here. You're providing true statements that are not answering the question. They are tangentially related, but I think you have missed exactly what is being asked for here. How does the polytope that you describe clearly represent $1^4+2^4+\cdots+n^4$? In the same way that the stairstep shape clearly represents $1+2+\cdots+n$? Mar 13, 2015 at 19:56
• And yes, of course $\sum k^2$ divides $\sum k^4$. All $\sum k^i$ are divisible by $n$ and $(n+1)$. And all $\sum k^{2i}$ are additionally divisible by $(2n+1)$. So not only does $\sum k^2$ divide $\sum k^4$, but also $\sum k^6$, $\sum k^8$, etc. I haven't seen a way to use that to help with my geometric question though. Mar 13, 2015 at 19:59