Larry Riddle, Agnes Scott College

Suppose the area of the original triangle **S(0)** is equal to
1. At the first iteration we remove (1/4)th of the area of
**S(0)**, so **S(1)** has area 3/4. Next we remove 3 triangles,
each having (1/4)th of the area of the triangle from which it is
taken, so the total area we remove is 3/16. This means that
**S(2)** has area 3/4−3/16 = 9/16 = (3/4)^{2}. It looks like
we should have that the area of **S(n)** is (3/4)^{n} for all
**n**. To see that this is indeed so, we can use induction. In
constructing **S(k+1)**, we remove 3^{k} triangles, each of
area (1/4)^{(k+1)}. Using our induction hypothesis, we get that
the area of **S(k+1)** would therefore be

$${\rm{area }} \: S(k + 1) = {\rm{ area }} \: S(k) - \frac{{{3^k}}}{{{4^{k + 1}}}} = \frac{{{3^k}}}{{{4^k}}} - \frac{{{3^k}}}{{{4^{k + 1}}}} = \frac{{4 \cdot {3^k} - {3^k}}}{{{4^{k + 1}}}} = \frac{3 \cdot {3^k}}{4^{k+1}} = {\left( {\frac{3}{4}} \right)^{k + 1}}$$

So we were right about the formula for the area of **S(n)** for
all **n**. Actually, there is a more direct way to see this. To
get **S(k+1)**, we scale **S(k)** by 1/2, which reduces the
area by 1/4 = (1/2)^{2}. But we make 3 copies of this scaled version to
form **S(k+1)**. Therefore the area of **S(k+1)** must be
(3/4)th of the area of **S(k)**. That's all we need to know to get
our formula since this implies by repeated backward application that the area of **S(n)** is equal to (3/4)^{n} of the area of **S(0)**.

Notice that the area of **S(n)** goes to 0 as **n** goes to infinity. This
means that we have removed "all" of the area of the original triangle
in constructing the Sierpinski gasket. But of course there are many
points still left in the gasket. That is one reason why area is not a
useful dimension for this set.