$\odot C_{1}$ is inscribed in a 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, to obtain a sequence of circles $C_{1}, C_{2}, C_{3}, \cdots$, find the sum of the areas of the first $n$ circles.
Assume $a>0$, by the symmetry of the circle and the parabola $y=a x^{2}+b x+c$ and the fact that the circle $C_{n}$ is internally tangent to the parabola, we can set the equation of the circle $C_{n}$ as: $\left(x+\frac{b}{2 a}\right)^{2}+\left(y-d_{n}\right)^{2}=r_{n}^{2}$, and solve the system of equations $\left\{\begin{array}{l}\left(x+\frac{b}{2 a}\right)^{2}+\left(y-d_{n}\right)^{2}=r_{n}^{2}, \\ y=a x^{2}+b x+c,\end{array}\right.$.
$$
y^{2}+\left(\frac{1}{a}-2 d_{n}\right) y+\left(d_{n}^{2}+\frac{b^{2}}{4 a^{2}}-\frac{c}{a}-r_{n}^{2}\right)=0,
$$

Since equation (1) has equal roots, then $\Delta=\left(\frac{1}{a}-2 d_{n}\right)^{2}-4\left(d_{n}^{2}+\frac{b^{2}}{4 a^{2}}-\frac{c}{a}-r_{n}^{2}\right)=0$, solving for $r_{n}^{2}=$ $\frac{b^{2}-1}{4 a^{2}}+\frac{1}{a}\left(d_{n}-c\right)$. Similarly, we get $r_{n+1}^{2}=\frac{b^{2}-1}{4 a^{2}}+\frac{1}{a}\left(d_{n+1}-c\right)$. Therefore, $r_{n+1}^{2}-r_{n}^{2}=$ $\frac{1}{a}\left(d_{n+1}-d_{n}\right)$.

Since the circle $C_{n}$ is externally tangent to the circle $C_{n+1}$, we have $r_{n+1}+r_{n}=\left|d_{n+1}-d_{n}\right|$. If $d_{n+1}2)$, there is only one circle $C_{1}$, which does not meet the problem's requirements. Thus, $d_{n+1}>d_{n}$, then $r_{n+1}+r_{n}=d_{n+1}-d_{n}, r_{n+1}-r_{n}=\frac{1}{a}$. Therefore, $\left\{r_{n}\right\}$ is an arithmetic sequence with the first term $r_{1}=\frac{1}{a}$ and common difference $\frac{1}{a}$, hence $r_{n}=\frac{n}{a}$, and the area of the circle $C_{n}$ is $\frac{\pi}{a^{2}} n^{2}$. Therefore, the sum of the areas of the first $n$ circles is: $\frac{\pi}{a^{2}}\left(1^{2}+2^{2}+\cdots+n^{2}\right)=\frac{\pi}{6 a^{2}} n(n+1)(2 n+1)$.