In a Koch's Snowflake- How do we derive the area of the 'n' th iteration?

Question

Falzoon has answered:
Area of n(th) snowflake = A * [8 - 3*(4/9)^n] / 5
But how does one derive this?
If there is any other method which does not consider the (n-1)th iteration, please show me how to derive those as well. THANK YOU

Answer 1

The following site shows how to get a formula for the area of the nth iteration:
http://mathworld.wolfram.com/KochSnowflake.html

On line 16 you have a formula that contains summation notation. The summation is the partial sum of a geometric series.

The partial sum of a geometric series is:
Sn = a1 (1-r^n) / (1-r)

See the following link for how to derive this:
http://www.richland.edu/james/lecture/m116/sequences/geometric.html

Use a1 =1 and r = 4/9. Then substitute Sn in equation 16 for the part containing the summation. Simplify it and you'll get Falzoon's formula.