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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.

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    +1. But I want no part in this madness...2012-11-21
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    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?2012-11-21
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    @StevenStadnicki Interesting!2012-11-21
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    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.2012-11-21
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    @Amzoti I'm not - I'll look for it at our local giant used book store.2012-11-21
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    Alex, you can actually peruse pages and see some of the proof strategies using Amazon. Here are some other choices to peruse: 1. Proofs without Words: Exercises in Visual Thinking (Classroom Resource Materials) (v. 1), 2. Proofs Without Words II: More Exercises in Visual Thinking (Classroom Resource Materials) (v. 2). I believe there are others too as I have seen some for topics such as Linear Algebra (IIRC). Enjoy!2012-11-21
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    30 comes from the fact that the Bernoulli number $B_4=-1/30$. See "Faulhaber's formula" in Wikipedia.2013-04-23
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    @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.)2013-04-23
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    I'm certainly gonna ask my teacher about this. He thinks he knows everything, and rules out answeres in many good books.2013-07-12
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    Here is a cute article on [Faulhaber's triangle](http://www.maa.org/sites/default/files/Torabi-Dashti-CMJ-2011.pdf)2015-03-12

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