Construct an infinite fractal. Stage 0 is a unit square. At each stage, a square is appended to the vertices of the previous stage such that the sides are 1/2 the sides of the previous stage and parallel to the sides. Find the area after infinite repetitions of this.
I know that this is a geometric series, but I am not sure how to find the common ratio.
On
Hints:
Each stage brings $3$ times as many vertices as the previous stage, so stage $n$ brings $4 \times 3^n$ for $n>0$
Each stage brings as many squares of area $\left(\frac{1}{2^n}\right)^2$ as there were new vertices at the previous stage, so stage $n$ brings an extra area of $4 \times 3^{n-1}\times \frac1{2^{2n}} = \left(\frac{3}4\right)^{n-1}$ when $n>0$
So the total area is $$1+\left(\frac{3}4\right)^{0}+\left(\frac{3}4\right)^{1}+\left(\frac{3}4\right)^{2}+\cdots$$ which, apart from the first term, is a geometric series
On
This appears to be the fractal under discussion, if it is meant that unoccupied vertices at each stage are to have half-scale squares appended to them. If that is the case, then it seems that "stage 1" does not belong to the geometric series either, so there will be one more term "outside the summation
$$ \mathbf{A} \ \ = \ \ 1 \ + \ \underbrace{4·\left(\frac{1}{4} \right)}_{\text{stage 1}} \ + \ 4 \ \sum_{k = 1}^{\infty} \ 3^k · \left(\frac{1}{4} \right)^{k+1} \ \ = \ \ 1 \ + \ 1 \ + \ \frac{4 · \frac{3}{16}}{1 - \frac{3}{4}} \ \ , $$
with "stage 1" having four vertices to add squares to, but all subsequent stages having only three "free" vertices to add "trees" to. (It is fairly common in self-similar constructions to have one or more terms that are not part of the principal geometric series. This may be what caused your uncertainty.) This sum agrees with the result given by Henry and Parcly Taxel . The factor of perimeter increase at each stage is $ \ \frac32 \ \ , $ so the geometric series for total perimeter of this figure diverges.
We can see from the diagram that the entire structure is bounded by a square nine times the area of the initial square. Since each stage is a decrease in linear scale by a factor of $ \ \frac12 \ \ , $ which has three times as many squares at that stage compared to the previous one, we can characterize this fractal by a "covering dimension"
$$ \mathcal{D} \ \ = \ \ - \ \frac{\log 3}{\log \frac12} \ \ \approx 1.585 \ \ , $$
as we would expect since this fractal incompletely "fills" the bounding square of dimension $ \ 2 \ . $
There is one big square (the initial one) and four copies of infinite trees starting from a square of side $\frac12$ and adding $\frac34$ of the area of the previous stage at each iteration. Thus the area of each tree is $$\frac{1/4}{1-3/4}=1$$ and the area of the whole fractal is $5$.