Prove by induction area of koch snowflake For all $n>=0$ prove that the area of a Koch snowflake is $a_n = a_0(\frac{8}{5}-\frac{3}{5}(\frac{4}{9})^n)$ where $a_0=\frac{\sqrt{3}}{4}$
I'm trying to get $P(n+1)$ from $P(n)$ but I'm not sure how to proceed. 
For $P(n)$ I now have $\frac{2\sqrt{3}}{5}-\frac{3\sqrt{3}}{20}(\frac{4}{9})^n$. 
The target is $P(n+1) = \frac{2\sqrt{3}}{5}-\frac{3\sqrt{3}}{20}(\frac{4}{9})^{n+1}$. 
 A: Do you have to use induction?
Let's say that the flake after $j$ steps has $n_j$ sides of length $s_j$, now since you replace each side with four of length $s_j/3$ and thereby adding a equilateral triangle of side $s_j$ you will end up with $s_{j+1}$ = $s_j/3$ and $n_{j+1} = 4n_j$ and the area added in this step to be $n_j a_0 s_j^2/9$.
Now it's easy to see from this that $s_j = s_03^{-j}$, $n_j = n_04^j$, so the area added is
$\sum a_0 n_j s_j^2/9 = a_0\sum n_0 4^j s_0^2 3^{-2j}/9 = a_0n_0s_0^2\sum(4/9)^j/9$
Then it's just a matter of putting it into the formula for geometric series:
$a_0n_0s_0^2{1 - (4/9)^{N+1}\over1-4/9}/9 = a_0n_0s_0^2{1-(4/9)^{N+1}\over 5/9}/9$
This is however only the area added to the original triangle that has to be added:
$A_n = a_0s_0^2 + a_0n_0s_0^2( {1-(4/9)^{N+1}\over 5/9})/9$
Now we start with a triangle so $n_0=3$ and with unit length we get:
$A_n = a_0( 1 + 3{1-(4/9)^{N+1}\over 5/9}/9) = a_0(1 + 3/5 - {3\over 5}(4/9)^{N+1}) = a_0({8\over5} - {3\over 5}({4\over 9})^{N+1})$
