Can you calculate $1+2+3+4+\ldots+k$ geometrically?

Question

I am learning Integrals from What is the sum $1+2+3+4+\ldots+k$? - Week $11$ - Lecture $2$ - Mooculus and I still don't get how to calculate $1+2+3+4+\ldots+k$ geometrically:

According to the video, $1+2+3+4+\ldots+k$ can be seen as a triangle like this: the triangle of $1+2+3+4+\ldots+k$

So I can make a copy of the triangle and combine them like this to get a rectangular: the rectangular, hence the area is $\frac{k(k+1)}{2}$.

But I realize that I don't have to assemble the rectangular, because the area of the triangle is easy to computer, which is $\frac{k\cdot k}{2}$.

So which one is the right one, which is one is the real geometric view of $1+2+3+4+\ldots+k$, $\frac{k^2}{2}$ or $\frac{k(k+1)}{2}$?

Answer 1

Actually, think of it not as a triangle, but as a trapezium (or a trapezoid if you are in the US).

The shape has a top length of $1$, not $0$, hence it is not a triangle.

You probably understand it but I will give you an example anyways: $$\sum_{n=1}^{5}n$$ ⬤
⬤⬤
⬤⬤⬤
⬤⬤⬤⬤
⬤⬤⬤⬤⬤

Let's double the dots to form a rectangle:

⬤◯◯◯◯◯
⬤⬤◯◯◯◯
⬤⬤⬤◯◯◯
⬤⬤⬤⬤◯◯
⬤⬤⬤⬤⬤◯

Number of black and white dots $=$ Length $\times$ Width

$$2\sum_{n=1}^{5}n=5\times(5+1)$$

The number of black dots is just half of it (We doubled it first)

$$\sum_{n=1}^{5}n=\frac{5\times(5+1)}{2}$$

To generalise it:

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

Answer 2

Thanks all for pointing out the whole picture is not a triangle, I focused too much on the picture made of dots, neglecting the fact that the number $1$ is a 'square', but not a dot, so combining them will actually get a stair.

I drew a new picture of $1+2+3+\ldots+k$.

When $k$ is large enough, the stair can be seen as a trapezoid

And the result of $1+2+3+4+\ldots+k$ can be seen as:

1. the area of the trapezoid: $(k+1)k/2$
2. half of the rectangular: $(k+1)k/2$

Wait! Is this how we get the formula for the area of a trapezoid??