Representing the areas of Sierpinski triangles as a partial sum of a geometric sequence? About Sierpinski triangles: Express the areas of the shaded triangles in the $n$th stage as a sum of a geometric series.
How do I represent the increasing triangles as a partial sum of a geometric series?
Sierpinski triangles: The Sierpinski triangle iterates an equilateral triangle (stage 0) by connecting the midpoints of the sides and shading the central triangle (stage 1). Repeat this process for the unshaded triangles in stage 1 to get stage 2.

 A: Counting the area of the white triangles:Let’s take this as an infinite geometric sequence. In our first cut, $\frac{x}{4}$ is removed, where $x$ is the original triangle’s area. This is indicated by the second triangle in the picture. In our second cut, $\frac{3x}{16}$ is removed, as indicated by the third triangle. In our third cut, $\frac{9x}{64}$ is removed, indicated by the fourth triangle. This pattern then continues on infinitely. Again, it is a sequence- $\frac{x}{4} , \frac{3x}{16}, \frac{9x}{64}......$ So in order to find the total amount of white space, we have to add all the terms here up. You might ask, well this is infinite, so how in the world do we add this up? Note that the first term is $\frac{x}{4}$ and the ratio is $\frac{3}{4}$. We now use the formula for an infinite geometric series:  

$$\frac{a_1}{1-r}$$ where $a_1$ is the first term of the geometric sequence and $r$ is its common ratio. So, on plugging the terms $a_1 = \frac{x}{4}$ and $r =\frac{3}{4}$ we have total white area equals to $\frac{\frac{x}{4}}{1-\frac{3}{4}} = x$. 

If you want an explicit formula for area of the white triangle in the $n$th figure starting from $n=2$, we have, area$_{n}$= $ \frac{x}{4}\times (\frac{3}{4})^{n}$ where for $n=1$, we have our area as $0$. 
Hope it helps.
