A couple weeks ago I posted a visual proof that

This says that the *n*th triangular number equals C(*n*+1, 2), the number of ways to choose two things from a set of *n* + 1 things.

I recently ran across a similar proof here. A simplex is a generalization of a triangle, and you can prove the equation above by counting the number of edges in simplices.

A 0-simplex is just a point. To make a 1-simplex, add another point and connect the two points with an edge. A 1-simplex is a line segment.

To make a 2-simplex, add a point not on the line segment and add two new edges, one to each vertex of the line segment. A 2-simplex is a triangle.

To make a 3-simplex, add point above the triangle and add three new edges, one to each vertex of the triangle. A 3-simplex is a tetrahedron.

Now proceed by analogy in higher dimensions. To make an an *n*-simplex, start with an *n*-1 simplex and add one new vertex and *n* new edges. This construction shows that the number of edges in an *n* simplex is 1 + 2 + 3 + … + *n*.

Another way to count edges is to note that an *n*-simplex has *n*+1 vertices and an edge between every pair of vertices. So an *n* simplex has C(*n*+1, 2) edges. So C(*n*+1, 2) must equal 1 + 2 + 3 + … + *n*.

## {{ parent.title || parent.header.title}}

## {{ parent.tldr }}

## {{ parent.linkDescription }}

{{ parent.urlSource.name }}