Snowflake By Haese Mathematics · Certified

This simplifies to: [ A_n = A_0 \left[ 1 + \sum_{k=1}^n \frac{3 \times 4^{k-1}}{9^k} \right] = A_0 \left[ 1 + \frac{1}{3} \sum_{k=1}^n \left(\frac{4}{9}\right)^{k-1} \right] ]

Repeat the process: take every straight line segment, divide it into three equal parts, and replace the middle third with two segments of that length. snowflake by haese mathematics

For each side, remove the middle third and replace it with two segments of the same length (forming an equilateral "bump"). The number of sides increases. This simplifies to: [ A_n = A_0 \left[

Using the sum of a geometric series with ratio ( r = \frac{4}{9} < 1 ), as ( n \to \infty ): [ A_{\infty} = A_0 \left[ 1 + \frac{1}{3} \times \frac{1}{1 - \frac{4}{9}} \right] = A_0 \left[ 1 + \frac{1}{3} \times \frac{9}{5} \right] = A_0 \left[ 1 + \frac{3}{5} \right] = \frac{8}{5} A_0 ] Using the sum of a geometric series with