2. Given that $\odot O_{0}$ is the incircle of the equilateral $\triangle A B C$, $\odot O_{1}$ is externally tangent to $\odot O_{0}$ and tangent to two sides of $\triangle A B C$, $\cdots, \odot O_{n+1}$ is externally tangent to $\odot O_{n}$ and tangent to two sides of $\triangle A B C$ $(n \in \mathbf{N})$. Then, the ratio of the sum of the areas of all possible circles (including $\odot O_{0}$, 3 $\left.\odot O_{n}(n \in \mathbf{N})\right)$ within $\triangle A B C$ to the area of $\triangle A B C$ is $\qquad$
2. $11 \pi: 24 \sqrt{3}$.

Let the radius of $\odot O_{n}$ be $R_{n}(n=0,1,2, \cdots)$.
Since the angle between the two external common tangents of $\odot O_{n}$ and $\odot O_{n+1}$ is $60^{\circ}$,
we have $\sin 30^{\circ}=\frac{R_{n}-R_{n+1}}{R_{n}+R_{n+1}}$,
which means $R_{n}=3 R_{n+1}, \frac{R_{n+1}}{R_{n}}=\frac{1}{3}$.
The total area of all circles is
$$
\begin{aligned}
S_{\text {total }} & =\pi R_{0}^{2}+3\left(\pi R_{1}^{2}+\pi R_{2}^{2}+\cdots\right) \\
& =\pi R_{0}^{2}+3 \cdot \frac{\pi R_{1}^{2}}{1-\frac{1}{9}} \\
& =\pi R_{0}^{2}+\frac{9}{2} \pi R_{1}^{2}=\frac{11}{8} \pi R_{0}^{2} .
\end{aligned}
$$

It is easy to find $S_{\triangle A B C}=3 \sqrt{3} R_{0}^{2}$.
$$
\therefore S_{\pi}: S_{\triangle A B C}=11 \pi: 24 \sqrt{3} \text {. }
$$