The number of red triangles at Stage n is 3n. 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 1/4n. So the total area of the red triangles at Stage n is (3/4)n.

The number of white triangles removed is 1 + 3 + 32 + ... 3n-1 and summing this geometric series the total number of triangles removed at Stage n is 1/2(3n - 1).

The total area of the triangles removed is given by the series:
( 1
4
 ) + 3( 1
4
 )2 + 32( 1
4
 )3 +... 3n-1( 1
4
 )n = 1 - ( 3
4
 )n.
and this is as expected from the calculation for the red triangles. As n ® ¥ 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 = md we have 3 = 2d so log3 = log(2d) and this gives
d = log 3
log 2
 » 1.585