Summation of Squares!!!

First, I started with a simple diagram with 5 adjacent squares:

I noticed that when I calculated the total area, I could start a big rectangle, with base $5 + 4 + 3 + 2 + 1$ and height $5$ . Then I could subtract many smaller rectangles. I noticed that as I summed the total extra area in each individual row, they made consecutive triangular numbers. Now I knew I could generalize this.

$1^{2}$ + $2^{2}$ + $3^{2}$ + $4^{2}$ + … + $n^{2}$ was going to start with a rectangle with base $(1 + 2 + 3 + 4 + … + n)$ and height $n$ . From there, I could subtract triangular numbers starting with $(n - 1) + (n - 2) + (n - 3) … + 1)$ , added to $(n - 2) + (n - 3) + (n - 4)$ and so on. $(n - 1)$ = $k$ .

The general formula for summation of triangular numbers from $1 \Rightarrow k$ is $\huge\frac{k \times {(k + 1)} \times {(k + 2)}}{6}$

So my overall formula was:

$\huge\frac{\color{#D61F06}{n^{2}} \color{#0C6AC7}{\times} \color{#20A900}{(n + 1)}}{\color{#EC7300}{2}} \color{teal}{-} \frac{\color{skyblue}{(n - 1)} \color{#0C6AC7}{\times} \color{hotpink}{(n + 1)} \color{#0C6AC7}{\times} \color{#69047E}{n}}{\color{#624F41}{6}}$

This can be simplified to the formula we know today:

$\huge \frac{n \times {(n + 1)} \times {(2n + 1)}}{6}$

Plug 3141 as $n$ , and we get 10334510871 .