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Koch Snowflake Proof by Induction. 
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#1
Jul1314, 08:06 AM

P: 11

Hi, I was wondering if there is a way to prove the area of the Koch Snowflake via induction?
At the moment I have the equations: A_{n+1}=A_{n}+[itex]\frac{3√3}{16}[/itex]([itex]\frac{4}{9}[/itex])^{n} and A_{n}=[itex]\frac{2√3}{5}[/itex][itex]\frac{3√3}{20}[/itex]([itex]\frac{4}{9}[/itex])^{n} These two don't seem to work together very well when trying to prove by induction. Can anyone offer any advice? This is not homework by the way :). 


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