Search for a geometrical way to calculate the summations $\sum_{k=1}^n k^p$ ($p=1,2,3...$)

Question

There exist geometrical proofs for the calculation of the sums $S_p(n):=\sum_{k=1}^nk^p$ for $p=1$ and $p=2$ (e.g. compilation of proofs for sum of squares and cubes, understanding some PWoWs).

The core question of this thread is: Does there exist a geometry-based method to proof the formulas of the above given summations, which is applicable for all integers $n$ and $k$?

I found geometrical proofs for $p=2$ and $p=3$, which I find beautiful. (I am sure other persons found these visualizations before, there are hints on this e.g. in the video Power sum master class.)

The method of this proof for $p=2$ consists in interpreting the sum $1^2+2^2+3^3+...$ as the volume of a shape built by square plates stacked on top of each other thus forming a kind of pyramid. The task is then to calculate the volume of this pseudo-pyramid. The method of the proof for $p=3$ is analogous, but now the 2d squares are replaced by 3d cubes and the 3d pseudo-pyramid is replaced by a 4d pseudo-pyramid. I will show these two proofs as answers to the question. I will also show a nice geometric proof for $p=1$ (which is also contained in the video mentioned above Power sum master class).

Remarks: I am convinced that I can generalize the method to $p=4$. And I am confident, that the method can be generalized to any number of $p$. I guess one has to find and write down quantities of and relations between the geometric features of the $(p+1)$-dimensional pyramids such as (hyper-)faces and sort out how many unit-hyper-cubes touch which faces. If the quantities and relations are written down in a systematic way, one can start generalizing the found rules in an algebraic way. Nevertheless the proof will still be of geometric kind, because the quantities will be connected to geometry in $(p+1)$ dimensions.

Answer 1

While showering, I found a geometrical proof for $p=2$, which can easily be transfered to $p=1$. I am convinced it can be generalized to $p=3$ - if one is familar with 4-dimensional geometry - and I am confident, that it can also be generalized to higher values of $p$ in $(p+1)$-dimensional geometry.

I will show the idea for the case $n=4$ and then generalize it to arbitrary $n$. Fig. 1 shows that the sum $1^2+2^2+3^2+4^2$ can be visualized by a shape in 3d, which is similar to a pyramid. Each square number of the sum is represented by a square which has been augmented to 3d by "pulling it up" by a length of 1. Hence each square number $i^2$ $(i=1, 2, 3, 4)$ is represented by a number of $i^2$ unit cubes, arranged in the pattern of a square. The desired sum corresponds to the volume of the 3d shape "pseudopyramid".

This volume can easily be approximated as $V_{pyr}=\frac{1}{3} \cdot base \cdot height=\frac{1}{3} \cdot n^2 \cdot n=\frac{1}{3}n^3=\frac{64}{3}=21\frac{1}{3}$. However, compared to the exact result $1^2+2^2+3^2+4^2=30$ something is missing. The missing volume is due to the parts of the cubes on the two front side-areas of the "pseudopyramid" which extend further than a similar real pyramide with base $4^2$ and height $4$ would do. The cubes on these areas are cut by these two side faces of the pyramid in parts.

In order to obtain the exact result one has to add these missing pieces. Some of the cubes are cut fifty-fifty into two pieces. These are marked in blue color in Fig. 2. Others are dissected by two side faces of the real pyramid such that inside the real pyramid only a rest pyramid of volume $\frac{1}{3}$ remains of each such unit cube. These are marked orange in Fig. 2. For the case $n=4$ one has to add $2 \cdot (1+2+3) \cdot \frac{1}{2}=6$ (blue) and $4 \cdot \frac{2}{3}=2\frac{2}{3}$ (orange). Hence, one arrives at the correct sum $21\frac{1}{3}+6+2\frac{2}{3}=30$.

This procedure can be generalized to the case $n$ to obtain the sum $S_n=1^2+2^2+3^2+...+n^2$. The volume of the real pyramid here is $V_{pyr}=\frac{1}{3} \cdot n^3$. The missing "blue" parts are $V_{blue}=2 \cdot (1+2+3+...+(n-1)) \cdot \frac{1}{2} = \frac{1}{2} (n-1) \cdot n$ (having applied the "Gauß-Formula"). The missing "orange" parts add up to $V_{orange}=n \cdot \frac{2}{3}$.

In total one obtains

$V_{pyr}+V_{blue}+V_{orange}=\frac{1}{3} \cdot n^3 + \frac{1}{2} (n-1) \cdot n + \frac{2}{3} \cdot n = \frac{1}{3} \cdot n^3 + \frac{1}{2} n^2 + \frac{1}{6} = \frac{1}{6}n(n+1)(2n+1)n$,

q.e.d.

Remark 1: I am quite sure that somebody else has already found this geometric view before, but so far I have not yet found it on math.stackexchange.com.

Remark 2: This proof is similar to https://math.stackexchange.com/a/2371512/930508. However, I think it is simpler and more straightforward and can be generalized easier.

Remark 3: At https://www.felixvoigt.de/wp-content/uploads/2023/07/summe_der_quadratzahlen_kompr.pdf you can find a more clear version of this geometric idea in German language.

Answer 2

There is an analogous geometrical view for the case $p=1$ in 2 dimensions. Look at Fig. 3, which illustrates the view for $n=4$. The pyramid in 3d corresponds to the triangle in 2d. The sum $1+2+3+4$ can be calculated as the area of the pseudo-triangle. One can approximate this area as the area of the triangle which has the dashed line as one side and afterwards add 4 half squares extending beyond the dashed line. $S_1(4) = \frac{1}{2} \cdot 4 \cdot 4 + 4 \cdot \frac{1}{2} = 8 + 2 = 10$. For arbitrary $n$ one arrives at the "Gauß-formula" $S_1(n) = \frac{1}{2} n \cdot n + \frac{n}{2} = \frac{1}{2} n \cdot (n+1)$. This geometric proof is also shown in the video Power sum master class.

Answer 3

Sum of cubic numbers, calculated in 4 dimensions

Recently I had an idea how to represent the 4th dimension. (Probably this is already well known, maybe I have also seen this somewhere before.) As the things you want to draw are normally restricted to a limited range, it is possible to use the $x$-axis a second time to represent the 4th dimension. If the 4-dimensional object is limited to the range [0, 1] in dimension 4, one just draws slices of the object at values 0, 0.25, 0.50, 0.75. That is sufficient, if the boundaries of the object vary in a linear way. (In a similar way one can also display 5 or 6 dimensional objects.)

A 4-dimensional cube would look the following way (Fig. 4). A 4-dimensional symmetric pyramid would look like this (Fig. 5). And a special case of an oblique 4d-pyramid looks like this (Fig. 6). This is an oblique pyramid, where the upper corner (in 4th dimension) is defined as $\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$.

The 4d “quasi-pyramid” which can be used to calculate the sum, looks this way (Fig. 7).

Analogously to the 3d-pyramid having 4 side-faces, which correspond to the 4 sides of a square, the 4d-pyramid has 6 side-volumes corresponding to the 6 faces of a cube. The little 4d-cubes inside the 4d-quasipyramid are cut only by those side-volumes of the 4d-pyramid, which do not include the point $\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$. There are 3 side-volumes, which include this point and 3 side-volume which don‘t. The ones which include the point are colored in red in Fig. 8 and the ones which do not in green. (The side-volumes extend in 4 dimensions, thus e.g. all four green-colored upper sides of the apparent four cubes in the figure belong to the same side-volume. Two red-colored side-volumes are not visible in the figure, also one green-colored one is hidden.)

Now, we paint the little cubes according to wether they are attached to one (blue), two (orange) or three (yellow) of the “green” side-volumes (Fig. 9).

The little 4d-cubes, corresponding to the blue-colored cubes are cut by a side-volume of the real pyramid into 2 equal parts. The orange-colored unit-4d-cubes are cut by 2 side-volumes such that 1/3 of a unit cube remains inside the real 4d-pyramid (to be proven). The yellow-colored cubes are cut by 3 side-volumes 3 times, such that inside the real pyramid only 1/4 of a unit cube remains, resembling the whole 4d-pyramid.

For the case $n=4$ the sum $S_3(4)=1^3+2^3+3^3+4^3$ can by approximated by $V_{pyr} = \frac{1}{4} base \cdot height = \frac{1}{4} 4^3 \cdot 4 = 64$ . In order to obtain the exact result, one has to add the missing pieces. These are

$V_{blue} = 3 \cdot ( 3^2+2^2+1^2 ) \cdot \frac{1}{2} = 3 \cdot 14 \cdot \frac{1}{2} = 21$,

$V_{orange} = 3 \cdot ( 3+2+1 )\cdot\frac{2}{3} = 3 \cdot 6 \cdot \frac{2}{3} = 12$,

$V_{yellow} = 1 \cdot ( 1+1+1+1 )\cdot\frac{3}{4} = \frac{9}{4} = 3$,

In total one obtains

$S_3(4) = 64 + 21 + 12 + 3 = 100$.

This is the correct result, as $1^3+2^3+3^3+4^3 = 1+8+27+64=100$.

We generalize this to the case of $n$ summands.

$V_{pyr} = \frac{1}{4} base \cdot height = \frac{1}{4} n^3 \cdot n = \frac{1}{4} n^4$,

$V_{blue} = 3 \cdot ( 1^2+2^2+3^2+...+[n-1]^2 ) \cdot \frac{1}{2} = \frac{3}{2} \cdot (\frac{1}{3} ([n-1]^3) + \frac{1}{2} [n-1]^2 + \frac{1}{6} [n-1] ) = \frac{1}{2} [n^3-3n^2+3n-1] + \frac{3}{4} [n^2-2n+1] + \frac{1}{4} [n-1] = \frac{1}{2} n^3 - \frac{3}{4} n^2 +\frac{1}{4} n$,

$V_{orange} = 3 \cdot (1+2+3+...+[n-1]) \cdot \frac{2}{3} = 2 \cdot {[n-1] \cdot \frac{n}{2}}=n^2-n$,

$V_{yellow} = 1 \cdot n \cdot \frac{3}{4} = \frac{3}{4} n$.

They sum up to

$S_3(n) = \frac{1}{4} n^4 + (\frac{1}{2} n^3 - \frac{3}{4} n^2 + \frac{1}{4} n) + (n^2-n) + \frac{3}{4} n = \frac{1}{4} n^4 + \frac{1}{2} n^3 + \frac{1}{4} n^2$,

q.e.d.

(A more readable form of this proof can be downloaded here).