The number of red triangles at Stage n is

3^{n}

. At each stage the red triangles remaining have area one quarter of the area of the red triangles at the previous stage so the area of each red triangle at Stage n is

{\frac{1}{4}}^{n}

. So the total area of the red triangles at Stage n is

(\frac{3}{4})^{n}

The number of white triangles removed is $1 + 3 + 3^{2} + . . . 3^{n - 1}$ and summing this geometric series the total number of triangles removed at Stage n is $\frac{1}{2} (3^{n} - 1)$ .

The total area of the triangles removed is given by the series:

$(\frac{1}{4}) + 3 (\frac{1}{4})^{2} + 3^{2} (\frac{1}{4})^{3} + . . . 3^{n - 1} (\frac{1}{4})^{n} = 1 - (\frac{3}{4})^{n} .$

and this is as expected from the calculation for the red triangles. As $n \to \infty$ the total area of the red triangles tends to 0 and the total area of the white triangles tends to 1.

Using the formula $n = m^{d}$ we have $3 = 2^{d}$ so $\log 3 = \log (2^{d})$ and this gives

$d = \frac{\log 3}{\log 2} \approx 1.585$