# Gauss enjoys MathJax

Mon, 06/05/2017 - 10:23

Recently, I added the MathJax on my website that allows me to write $\LaTeX$ formulas and I couldn't help posting something that involves Math, yet at the same time keeping it simple for a wide range of audience. So just let me start by a motivational story.

According to the story, Carl Friedrich Gauss, in school, at the age of 8, was able to add the first 100 natural numbers using a special trick he came up with in a matter of minutes, in that way catching his teacher by surprise. I don't know if the story is true, but just imagine if you were a teacher, why would you ever give your 8 y/o pupils a task to add all the numbers from 1 to 100 during the class? Perhaps the teacher wanted to keep his/her class busy, while he/she can occupy him/her-self by doing something else. Napping could be one of the options I guess if you expect them to calculate the sum in the pedestrian way. Apparently, Gauss was not an ordinary kid to consider the straight and inverted orders of that summation.

$$^{ ^{\LARGE +}} \begin{matrix} 1 & + & 2 & + & 3 & + & \dots & + & 98 & + & 99 & + & 100 \\ 100 & + & 99 & + & 98 & + & \dots & + & 3 & + & 2 & + & 1 \\ \hline 101 & + & 101 & + & 101 & + & \dots & + & 101 & + & 101 & + & 101 \end{matrix} \tag{1}$$

or $2S_{100}=100\times101$ where $S_{100}=5050$ is the sum of first 100 integer numbers.

But there is nothing special about 100. The same way for $\forall n \in \mathbb N$ one can get

$$\sum_{i=1}^n i=\frac{n(n+1)}{2} \tag{2}$$

So far so good, but what would Gauss do if the teacher asked him to calculate the sum of squares ?

$$\sum_{i=1}^n i^2=1^2+2^2+3^2+\dots+n^2 \tag{3}$$

Gauss was smart, so he would have written the numbers in equilateral triangles in three different configurations.  $\def\iddots{ {\kern0mu\raise0mu{.}\kern0mu\raise3.9mu{.}\kern0mu\raise8.1mu{.}}}$

$$\begin{matrix} & & & & & 1 & & & & & \\ & & & & 2 & & 2 & & & & \\ & & & 3 & & 3 & & 3 & & & \\ & & 4 & & 4 & & 4 & & 4 & & \\ & \iddots & & & & & & & & \ddots & \\ n & & & & & \dots & & & & & n \end{matrix}$$

$$+$$

$$\begin{matrix} & & & & & n & & & & & \\ & & & & \iddots & & & & & & \\ & & & 4 & & & & \ddots & & & \\ & & 3 & & 4 & & & & & & \\ & 2 & & 3 & & 4 & & & & & \\ 1 & & 2 & & 3 & & 4 & & \dots & & n \end{matrix}$$

$$+\tag{4}$$

$$\begin{matrix} & & & & & n & & & & & \\ & & & & & & \ddots & & & & \\ & & & \iddots & & & & 4 & & & \\ & & & & & & 4 & & 3 & & \\ & & & & & 4 & & 3 & & 2 & \\ n & & \dots & & 4 & & 3 & & 2 & & 1 \end{matrix}$$

$$||$$

$$\begin{matrix} & & & & & 2n+1 & & & & & \\ & & & & & \vdots & & & & & \\ & & & & & & & & & & \\ & & & & & 2n+1 & & & & & \\ & \iddots & & & & & & & & \ddots & \\ 2n+1 & & & & & \dots & & & & & 2n+1 \end{matrix}$$

or

$$3(1^2+2^2+3^2+4^2+\dots+n^2)=(1+2+3+4+\dots+n)(2n+1) \tag{5}$$

That reduces the problem to eq. (2) the initial problem of Gauss. So plugging (2) in (5) Gauss would have got

$$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} \tag{6}$$

Oh, merciless teacher and poor Gauss! What if he was told to add a sequence of cubes $\sum_{i=1}^n i^3$? Have I already told that Gauss was smart? Gauss would have been smart enough to notice that each $i^{th}$ term of the sum can be represented by $i$ squares of size $i \times i$ glued side by side along a straight line to make a rectangle of size $i \times i^2$ and having an area of $i^3$ . He would have noticed also that the side of $i-1$ squares of size $i \times i$ can completely cover the side of $i$ squares of size $(i-1) \times (i-1)$, because $(i-1)i=i(i-1)$ makes an identity. This last fact allows to draw the following square of size $n(n+1) \times n(n+1)$ covered by nested belts of decreasing thickness by 1 at a time.

Inside each square belt of thickness $i$ Gauss can count $4i$ squares of size $i \times i$. So $i^{th}$ square belt sweeps out an area of $4 \times i \times i^2$. If Gauss summed up all the belt areas he would have obtained the area of the square of size $n(n+1) \times n(n+1)$.

$$4\times 1 \times 1^2 + 4\times 2 \times 2^2 + 4\times 3 \times 3^2 + \dots + 4\times n \times n^2 = 4 \times \sum_{i=1}^n i^3 = n^2 (n+1)^2 \tag{7}$$

This is beautiful beyond belief, because if we take into account (2) we can write $$\sum_{i=1}^n i^3={\left(\sum_{i=1}^n i\right)}^2 \tag{8}$$

Now let me challenge you by asking if the generalized trick of visualisation exists for $\forall p$ in Faulhaber's formula

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