9、 (16 points) As shown in the figure, there are $2^{n}$ points on the parabola $y^{2}=2 p x(p>0)$, denoted as $P_{1}, P_{2}, \ldots, P_{2^{n}}$, and $F$ is the focus of the parabola. The angles between the lines connecting each adjacent pair of points and $F$ are equal, i.e.,
$$
\angle P_{1} F P_{2}=\angle P_{2} F P_{3}=\cdots=\angle P_{2^{n}} F P_{1}
$$

Let the distance from $P_{i}$ to the directrix be $d_{i}$. Find
$$
\sum_{i=1}^{2^{n}} \frac{1}{d_{i}}
$$
Solution:
As shown in the figure, let $F P_{1}$ make an angle $\alpha$ with the positive x-axis.
From the problem, we know
$$
\begin{array}{c}
\angle P_{1} F P_{2}=\angle P_{2} F P_{3}=\cdots=\angle P_{2^{n}} F P_{1}=\frac{2 \pi}{2^{n}} \\
d_{i}=\left|P_{i} P^{\prime}\right|=\left|F P_{i}\right|=|F G|+\left|F P_{i}\right| \cos \left((i-1) \frac{\pi}{2^{n-1}}+\alpha\right) \\
=p+d_{i} \cos \left((i-1) \frac{\pi}{2^{n-1}}+\alpha\right)
\end{array}
$$

Therefore,
$$
d_{i}=\frac{p}{1-\cos \left((i-1) \frac{\pi}{2^{i-1}}+\alpha\right)}(i=1,2, \cdots, n)
$$
(8 points)
Then
$$
\sum_{i=1}^{2^{n}} \frac{1}{d_{i}}=\frac{1}{p} \sum_{i=1}^{2^{n}}\left(1-\cos \left(\frac{i-1}{2^{n-1}} \pi+\alpha\right)\right)=\frac{2^{n}}{p}-\frac{1}{p} \sum_{i=1}^{2^{n}} \cos \left(\frac{i-1}{2^{n-1}} \pi+\alpha\right)=\frac{2^{n}}{p}
$$