Given the parabola equation as $y=a x^{2}+b x+c(a \neq 0)$. Inside it, construct a circle $\odot C_{1}$ with radius $\frac{1}{|a|}$ that is internally tangent to the parabola. Construct $\odot C_{2}$ to be externally tangent to $\odot C_{1}$ and internally tangent to the parabola. Construct $\odot C_{3}$ to be externally tangent to $\odot C_{2}$ and internally tangent to the parabola, and so on, resulting in a sequence of circles $C_{1}, C_{2}, C_{3}, \cdots$, $C_{n}, \cdots$. Find the sum of the areas of the first $n$ circles.
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\text{I. The center coordinates of } C_{n} \text{ are } \left(-\frac{b}{2 a}, d_{n}\right), \text{ with radius } r_{n} (n=1,2, \cdots). \text{ Then the equation of the circle } C_{n} \text{ is: } \left(x+\frac{b}{2 a}\right)^{2}+\left(y-d_{n}\right)^{2}=r_{n}^{2}, \text{ combined with } y=a x^{2}+b x+c, \text{ we get } a y^{2}+\left(1-2 a d_{n}\right) y+\frac{b^{2}}{4 a}-c+a d_{n}^{2}-a r_{n}^{2}=0. \text{ Since the circle and the parabola are internally tangent, the quadratic equation in } y \text{ has equal roots } \Rightarrow\left(1-2 a d_{n}\right)^{2}-4 a\left(\frac{b^{2}}{4 a}-c+a d_{n}^{2}-a r_{n}^{2}\right)=0. \\
\text{Simplifying, we get } 4 a^{2} r_{n}^{2}=4 a d_{n}+b^{2}-4 a c-1. \\
\text{Similarly, for the circle } C_{n-1}, \text{ we have } 4 a^{2} r_{n-1}^{2}=4 a d_{n-1}+b^{2}-4 a c-1. \\
\text{By (1)-(2), we have } a\left(r_{n}^{2}-r_{n-1}^{2}\right)=d_{n}-d_{n-1}. \\
\text{Since } C_{n} \text{ and } C_{n-1} \text{ are externally tangent, we get } r_{n}+r_{n-1}=\left|d_{n}-d_{n-1}\right| \Rightarrow r_{n}-r_{n-1}=\frac{1}{|a|}, r_{1}=\frac{1}{|a|} \Rightarrow \left\{r_{n}\right\} \text{ is an arithmetic sequence with common difference } \frac{1}{|a|}. \\
\therefore r_{n}=\frac{n}{|a|} \cdot S_{n}=\pi r_{n}^{2}=\frac{\pi}{a^{2}} n^{2} . \\
\begin{aligned}
\therefore S & =S_{1}+S_{2}+\cdots+S_{n} \\
& =\frac{\pi}{a^{2}}\left(1^{2}+2^{2}+\cdots+n^{2}\right) \\
& =\frac{\pi}{6 a^{2}} n(n+1)(2 n+1) .
\end{aligned}
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