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John Cook is an applied mathematician working in Houston, Texas. His career has been a blend of research, software development, consulting, and management. John is a DZone MVB and is not an employee of DZone and has posted 172 posts at DZone. You can read more from them at their website. View Full User Profile

Triangular Numbers and Simplices

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A couple weeks ago I posted a visual proof that

1 + 2 + 3 + \cdots + n = {n+1 \choose 2}

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 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.

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