Can you calculate $1+2+3+4+\ldots+k$ geometrically? 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 $\frac{k(k+1)}{2}$?
 A: 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:
$$\sum_{n=1}^{k}n=\frac{k(k+1)}{2}$$
A: 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:

*

*the area of the trapezoid: $(k+1)k/2$

*half of the rectangular: $(k+1)k/2$
Wait! Is this how we get the formula for the area of a trapezoid??
