Finding the area of the Koch snowflake with geometric series

[Euphoriet]

I'm trying to find the area for the Koch Snowflake... I can't really figure it out... I found a couple of equations online (I can't really understand how most work.. or how they have been "derived".. including the mathworld one)

I'm looking at the following formula (seems to work.. but I can't figure out the (1-(4/9)^n)/(1-4/9) part.. something to do with an infinite geometric series right?
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Equation:

(sqr(3))/4) + [ ((sqr(3)/12 )) * (1-(4/9)^s)) ] / (1-4/9)

(sorry I could not figure out the pretty print stuff either.)
s = stage of snowflake/fractal, starting from 0

 

[Euphoriet]

oops.. sorry about the other thread.. oh well

 

[HallsofIvy]

I'm trying to find the area for the Koch Snowflake... I can't really figure it out... I found a couple of equations online (I can't really understand how most work.. or how they have been "derived".. including the mathworld one)

I'm looking at the following formula (seems to work.. but I can't figure out the (1-(4/9)^n)/(1-4/9) part.. something to do with an infinite geometric series right?

Yes, that is the sum, from i= 0 to n-1, of (4/9)^i

---
Equation:

(sqr(3))/4) + [ ((sqr(3)/12 )) * (1-(4/9)^s)) ] / (1-4/9)

(sorry I could not figure out the pretty print stuff either.)
s = stage of snowflake/fractal, starting from 0

Uh, that's not an equation! There is no "=". I'm not sure what you want to do with it. Add the two fractions? If so, then, of course, 1- 4/9= 5/9 so "(1-(4/9)^s/(1- 4/9)" is
$$\frac{1-\left(\frac{4}{9}\right)^s}{1- \frac{4}{9}}= \frac{1-\left(\frac{4}{9}\right)^s}{5/9}= \frac{9(1-\left(\frac{4}{9}\right)^s)}{4}$$
I come out with, for the whole thing,
$$\frac{\sqrt{3}+ \frac{3\sqrt{3}}{4}\left(1- \frac{4}{9}\right)^s}{4}$$

Double click on the "pretty print" (Latex) to see the code.