I'm working on a proof that looks like this:
Let $n$ be a positive integer. Given an equilateral triangle, place $n$ points on each side, dividing the side into $n+1$ equal segments.
Use the points to draw $n$ line segments parallel to each side of the triangle ($3n$ line segments in all).
Prove by induction that this will always divide the triangle into exactly $(n+1)^2$ little equilateral triangles.
But I am unsure of how to proceed past the base case.
Best Answer
From a triangle with $(n+1)$-fold subdivision you can remove $n+1$ small triangles that have an edge on the base line and the $n$ triangles between them that have a vertex on the base line. The result is a triangle with $n$-fold subdivision. Thus if $f(n)$ denotes the number of small triangles, we have $$f(n+1)=f(n)+(n+1)+n.$$