Sierpinski Counter

We begin with one triangle, to which we add a smaller triangle in the middle. After this, there are 3 spaces where we can add even smaller triangles, and then $3^2$ spaces, and then $3^3$ spaces, and so on. Every time we add a new, smaller triangle to the previous iteration, we gain 4 new triangle shapes (look at the upper middle image if this is unclear). Thus, the expression for the number of triangles in a partially-completed Sierpinksi triangle after $n$ such iterations is equal to $1+\sum^{n-1}_{n=0}4(3^n)$ (where $n=0$ for the first empty triangle). A classic formula for such an exponential sum is $\sum^{n-1}_{n=0}=\frac{1-r^n}{1-r}$ . Accordingly, our sum becomes $1+4(\frac{1-3^n}{-2})$ , which cancels down to $2(3^n)-1$ . The last triangle represents $n=8$ , so plugging 8 into our formula yields $2(3^8)-1=\color{#20A900}{\boxed{13121}}$