This note is one of many taken during 2026

Geometric Integer Sums of Lines and Squares

August 6th 2026

On the sums of natural numbers

\text{Let } L_n = 1 + 2 + 3 + \ldots + n = ?

Think about each integer as an area, and draw each component integer of $L_n$ ascending, in a square of dimension $n$ :

In this case, $n=4$ and $L_4 = 10$ .

Notice the diagonal set of squares from the bottom left to the top right of the greater square. There exist $n$ units on this diagonal. Above the diagonal, there are zero units as part of the greater square. Below the diagonal, all units are part of the greater square.

We know the diagonal has a total area of $n$ , but what about the area under the diagonal? If we consider the greater square with the diagonal removed, then this under-area is clearly half of the greater square’s (sans diagonal) area.

Thus the area of the under-area is $\frac{n^2 -n}{2}$ units

To find $L_4$ , the diagonal must be added back

L_4 = \frac{n^2 - n}{2} + n = \frac{n^2 + n}{2}

Proof By Induction

Base condition:

L_1 = \frac{1+1}{2} = 1

Following values

L_{n+1} = L_{n} + n + 1

\frac{(n+1)^2 + n}{2} = \frac{n^2+n}{2} + n + 1

Expand left side

\frac{n^2+2n+1 + n + 1}{2} = \frac{n^2+n}{2} + n + 1

Both sides are equal.

On the sums of squares of natural numbers

\text{Let } S_n = 1 + 2^2 + 3^2 + \ldots + n^2 = ?

Or recursively:

S_n = S_{n-1} + n^2

(Note, this section follows from a YouTube video by Perfect Yarn)

Thing about each squared integer as a square. Arrange them ascending:

In this case, $n = 3$ and $S_3 = 14$ .

Just as with the un-squared natural numbers, a closed form for $S_n$ can be found by interpreting this geometric construction.

If a equation for the quantity of units in each row $R_{j}$ is found, then the total $S_n$ is the sum of the row $R_{j}$ values. Where $j$ is the row number, from bottom up, starting at $0$ :

S_n = R_{0} + R_{1} + \ldots + R_{n-1}

For the bottom row ( $j=0$ ), $R_{0,3} = L_3$ (where $L_n$ is the sum of natural numbers from 1…n) or

R_{0,3} = \frac{3^2+3}{2}

For the following values of $j > 0$ , $R_{j}$ will decrease. Each increase of $j$ will cause a subsequent quantity to be removed:

R_{j} = \frac{n^2+n}{2} - \frac{j^2+j}{2}

Again, recall that

S_n = R_{0} + R_{1} + \ldots + R_{n-1}

Substituting in, $R_{j}$ we see:

S_n = n\left(\frac{n^2+n}{2}\right) - \frac{0^2+0}{2} - \frac{1^2+1}{2} - \ldots - \frac{(n-1)^2+(n-1)}{2}

Factor

S_n = \frac{1}{2}\left(n\left({n^2+n}\right) - (0^2 + 1^2 + \ldots + (n-1)^2) - (0 + 1 + \ldots + (n-1))\right)

The sum of natural numbers of the right of the equation can be simplified to $L_{n-1}$ and the sum of squares on the left can be simplified to $S_{n-1}$

S_n = \frac{1}{2}\left(n\left({n^2+n}\right) - S_{n-1} - \frac{(n-1)^2+(n-1)}{2}\right)

Recall that:

\begin{aligned} S_n &= S_{n-1} + n^2\\ S_{n-1} &= n^2 - S_n\\ \end{aligned}

Substitute $S_{n-1}$

S_n = \frac{1}{2}\left(n\left({n^2+n}\right) - S_n + n^2 - \frac{(n-1)^2+(n-1)}{2}\right)

Factor $S_n$ to one side

S_n = \frac{1}{3}\left(n\left({n^2+n}\right) + n^2 - \frac{(n-1)^2+(n-1)}{2}\right)

Simplify

S_n = \frac{1}{3}n\left(n^2+\frac{3}{2}n + \frac{1}{2}\right)

S_n = \frac{1}{6}n\left(2n^2+3n + 1\right)

Proof By Induction

Base condition

S_1 = \frac{1}{6}\left(2\cdot 1^2 +3\cdot 1 + 1\right) = \frac{6}{6} = 1

Following values

S_{n+1} = S_n + (n+1)^2

\frac{1}{6}(n+1)\left(2(n+1)^2+3(n+1) + 1\right) = \frac{1}{6}n\left(2n^2+3n + 1\right) + (n+1)^2

Simplify

(n+1)\left(2n^2+7n+6\right) = n\left(2n^2+3n + 1\right) + 6n^2+12n+6

2n^3+9n^2+13n+6 = 2n^3+9n^2 +13n+ 6

Both sides are equal.

This note is one of many taken during 2026