Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
+1. But I want no part in this madness... –  Benjamin Dickman Nov 21 '12 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? –  Steven Stadnicki Nov 21 '12 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. –  Amzoti Nov 21 '12 at 18:28
30 comes from the fact that the Bernoulli number $B_4=-1/30$. See "Faulhaber's formula" in Wikipedia. –  pharmine Apr 23 '13 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.) –  alex.jordan Apr 23 '13 at 4:16

2 Answers 2

On this book:

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 $$\sum_{i=1}^ni^4$$

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

share|cite|improve this answer
+1 This is cool, but I'm holding out for five-dimensional blocks :) –  alex.jordan Oct 21 '13 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 $L(\text{int}(P), t)= (-1)^d L(P,-t)$.

share|cite|improve this answer
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. –  alex.jordan Mar 12 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? –  Grigory M Mar 12 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. –  john mangual Mar 12 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 $1+2+\cdots+n$? –  alex.jordan Mar 13 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. –  alex.jordan Mar 13 at 19:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.