Sierpinski's Triangle - 4

After the first stage, the white triangle takes up $\frac{1}{4}$ of the total figure, and the red triangles take up $\frac{3}{4}$ of it.

After the second stage, the new white triangles take up $\frac{1}{4}$ of the old red triangles, which took up $\frac{3}{4}$ of the total figure, so they take up $\frac{1}{4}\cdot\frac{3}{4}$ of the total figure. Similarly, the red triangles now take up $\big(\frac{3}{4}\big)^2$ of the total figure.

After this, it should be clear that this pattern persists. At the $n$ th stage, the new white triangles take up $\frac{1}{4}\cdot\big(\frac{3}{4}\big)^{n-1}$ of the total figure, and the red triangles take up $\big(\frac{3}{4}\big)^{n}$ of the total figure.

This actually gives us two different ways to think about the area of the white triangles:

1. Because the total area is 1, however much of the total area the white triangles take up (as $n\rightarrow\infty$ ) is just their area

2. Similarly, we can subtract how much the red triangles will take up (as $n\rightarrow\infty$ ) from the total area (which is 1) to find the area of the white triangles

Method 1: Adding up the area of the new white triangles at each stage and taking the limit gives $lim_{n\rightarrow\infty}{\sum_{k=1}^{n}{\frac{1}{4}\big(\frac{3}{4}\big)^{k-1}}}={\sum_{k=1}^{\infty}{\frac{1}{4}\big(\frac{3}{4}\big)^{k-1}}}=\frac{1}{4}{\sum_{k=1}^{\infty}{\big(\frac{3}{4}\big)^{k-1}}}=\frac{1}{4}\sum_{k=0}^{\infty}{\big(\frac{3}{4}\big)^{k}}=\frac{1}{4}\cdot\frac{1}{1-\frac{3}{4}}=\frac{1}{4}\cdot\frac{1}{\frac{1}{4}}=\frac{1}{4}\cdot\frac{4}{1}=1$

Method 2: Taking the limit of the area of the red triangles gives $lim_{n\rightarrow\infty}\big(\frac{3}{4}\big)^{n}=0$ Obviously, $1-0=1$