Koch's Snowflake & Antisnowflake: Find Finite Area Formula

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The discussion focuses on deriving the finite area formulas for Koch's snowflake and antisnowflake. The area for Koch's snowflake is established as 81x(8/5), while the area for the antisnowflake is 81x(2/5). A recurrence relation is utilized to express the area of the snowflake, defined as A_n = A_n-1 + 27*(4/9)^(n-1), which converges as n approaches infinity due to the geometric series with a ratio of 4/9.

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I have a question to do with Koch's snowflake and anti snowflake..
I have no clue how to find the formula for the total area of each.
This is my table for snowflake
http://img485.imageshack.us/img485/8959/snowflake2an.jpg
& this is it for anti-snowflake
http://img356.imageshack.us/img356/2616/antisnowflake8vz.jpg

I've realized that the area for both are finite, with the snowflake being 81x(8/5) and antisnowflake being 81x(2/5)
 
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since you know how much area is added, you can set up a recurrence relation. for the snowflake, A_n = A_n-1 + 27*(4/9)^(n-1). since A_n-1 = An-2 + 27*(4/9)^(n-2) and so forth, A_n = A_0 + 27*(4/9)^(1-1) + 27*(4/9)^(2-1) + ... + 27*(4/9)^(n-1). this is a geometric series with the ratio 4/9<1 so it will converge as n goes to infinity.
 
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