5. As shown in the figure, divide an equilateral triangle with a side length of 1 into four congruent equilateral triangles, and remove the middle small triangle for the first time. Then, from the remaining three small equilateral triangles, remove the middle small triangle from each, keeping their sides, and repeat this process. The resulting set is the Sierpinski sieve.

Let $A_{n}$ be the sum of the areas of the small triangles removed at the $n$-th step (for example, $A_{1}$ is the area of the middle small triangle removed at the first step, and $A_{2}$ is the sum of the areas of the three small triangles removed at the second step), then the sum of the areas of all the small triangles removed in the first $n$ steps is
Answer $\frac{\sqrt{3}}{4}\left(1-\left(\frac{3}{4}\right)^{n}\right)$.
Analysis The original equilateral triangle area is $\frac{\sqrt{3}}{4}$, and in the $k$-th step, a total of $3^{k-1}$ small triangles are removed, $A_{k}=\frac{\sqrt{3}}{16}\left(\frac{3}{4}\right)^{k-1}$, therefore, the sum of a geometric series formula can be used, to get the answer as:
$$
\sum_{k=1}^{n} A_{k}=\frac{\sqrt{3}}{16} \sum_{k=1}^{n}\left(\frac{3}{4}\right)^{k-1}=\frac{\sqrt{3}}{16} \frac{1-\left(\frac{3}{4}\right)^{n}}{1-\frac{3}{4}}=\frac{\sqrt{3}}{4}\left(1-\left(\frac{3}{4}\right)^{n}\right)
$$