A couple weeks ago I posted a visual proof that
This says that the nth 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 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.