Recursive sequence for the Koch Snowflake 

Let $C_0$ an equilateral triangle of side 1 and $A_{n}$ the area of
the figure $C_{n+1}, n\in\mathbb{N}\cup {0}$
Write $A_{n+1}$ based on $A_n$ and the area of $C_0$

I noted that $C_1$ has been built by adding a smaller equilateral triangle on each side of $C_0$. Then $C_2$ is built by adding an even smaller equilateral triangle on each side of $C_1$. Figure $C_3$ is constructed by adding a smaller equilateral triangle on each side of $C_2$. So each new figure is formed by adding small equilateral triangles on the sides of the previous figure.
The area of one smaller triangle added in each figure can be calculated using the formula:
$$a=\frac{s^2\sqrt{3}}{4}$$
s is the length of the side, and knowing that the side of $C_0$ is $1$, I noted that the length of each segment of $C_{n+1}$ is given by $\left(\frac{1}{3}\right)^{n+1}$ so
$$a=\frac{\left(\left(\frac{1}{3}\right)^{n+1}\right)^2\sqrt{3}}{4}$$
I also noted that the number of sides in figure $C_{n+1}$ is $3 \times 4 ^ n$ so:
$$A_{n+1}=A_n+\frac{\left(\left(\frac{1}{3}\right)^{n+1}\right)^2\sqrt{3}}{4}\times 3\times 4^n=A_n+\left(\frac{1}{9}\right)^{n+1}\times \frac{\sqrt{3}}{4}\times3\times4^n$$
$$\boldsymbol{A_{n+1}=A_n+\frac{3\sqrt{3}}{16}\left(\frac{4}{9}\right)^{n+1}}$$

Is there another way or  reasoning to get the same recursive sequence?

One alternative reasoning might follow from:
$$A_{n+2} - A_{n+1} = \frac{4}{9} \left(A_{n + 1} - A_{n}\right) \quad;\quad A_1 = \frac{4}{3} A_0 \quad;\quad A_0 = \frac{\sqrt{3}}{4}\\
A_{n+2} - A_0 = \left(A_{n+2} - A_{n+1}\right) + \left(A_{n+1} - A_n\right) + \cdots + \left(A_1 - A_0\right)$$
I'll leave you to flesh out the details.
