Infinite Series -1

The external triangle has an area of 1.

The next smaller triangle has an area of $\frac{1}{4}$ , since the drawn lines divide the original triangle into four equal triangles.

The next smaller triangle now has $\frac{1}{4}$ of the area of the sorrounding triangle. In total its area is: $A = \frac{1}{4}*\frac{1}{4} = \frac{1}{4^{2}}$

This pattern continues, so that we can write the total area as an infinite sum: $A = 1 + \frac{1}{4} + \frac{1}{4^{2}} + \frac{1}{4^{3}}...$

Let's multiply the equasion by 4: $4A = 4* (1 + \frac{1}{4} + \frac{1}{4^{2}} + \frac{1}{4^{3}}...)$ $4A = 4 + 1 + \frac{1}{4} + \frac{1}{4^{2}}...$

We see that the term: $1 + \frac{1}{4} + \frac{1}{4^{2}}...$ , which is our original A , shows up again. That's why we can write:

$4A = 4 + A$ $3A = 4$ $A = \boxed{\frac{4}{3}}$