Question 5 Calculate: $\sum_{k=0}^{2017} \frac{5+\cos \frac{\pi k}{1009}}{26+10 \cos \frac{\pi k}{1009}}$. [4] (2018, Carnegie Mellon Informatics and Mathematics Competition)
【Analysis】Observing the connection between the numbers 2017 and 1009, we can transform $\cos \frac{\pi k}{1009}$ into $\cos \frac{2 \pi k}{2018}$.
Then $1, \cos \frac{2 \pi}{2018}+\mathrm{i} \sin \frac{2 \pi}{2018}$,
$$
\begin{array}{l}
\cos \frac{4 \pi}{2018}+\mathrm{i} \sin \frac{4 \pi}{2018}, \cdots, \\
\cos \frac{2 \times 2017 \pi}{2018}+\mathrm{i} \sin \frac{2 \times 2017 \pi}{2018}
\end{array}
$$

are the roots of the polynomial $P(x)=z^{2018}-1$.
By Euler's formula:
$$
\begin{array}{l}
\mathrm{e}^{\mathrm{i} \theta}=\cos \theta+\mathrm{i} \sin \theta, \\
\mathrm{e}^{\mathrm{i} \theta}=\cos \theta-\mathrm{i} \sin \theta,
\end{array}
$$

we get $\cos \theta=\frac{\mathrm{e}^{\mathrm{i} \theta}+\mathrm{e}^{-\mathrm{i} \theta}}{2}$.
In this problem, $\cos \frac{2 \pi k}{2018}=\frac{\mathrm{e}^{\frac{2 \pi k}{2018} \mathrm{i}}+\mathrm{e}^{-\frac{2 \pi k}{2018} \mathrm{i}}}{2}$.
Let $z_{k}=\mathrm{e}^{\frac{2 \pi k}{2018} \mathrm{i}}$. Then
$$
\cos \frac{2 \pi k}{2018}=\frac{z_{k}+z_{k}^{-1}}{2}(k=0,1, \cdots, 2017).
$$

From the above analysis, we know that $z_{k}$ are the roots of the polynomial
$$
P(x)=z^{2018}-1.
$$

Hence
$$
\begin{array}{l}
\sum_{k=0}^{2017} \frac{5+\cos \frac{\pi k}{1009}}{26+10 \cos \frac{\pi k}{1009}} \\
=\sum_{k=0}^{2017} \frac{5+\frac{z_{k}+z_{k}^{-1}}{2}}{26+10 \times \frac{z_{k}+z_{k}^{-1}}{2}} \\
=\sum_{k=0}^{2017} \frac{z_{k}^{2}+10 z_{k}+1}{2\left(z_{k}+5\right)\left(5 z_{k}+1\right)} \\
=\sum_{k=0}^{2017} \frac{1}{10}+\frac{1}{2} \sum_{k=0}^{2017} \frac{1}{z_{k}+5}-\frac{1}{50} \sum_{k=0}^{2017} \frac{1}{z_{k}+\frac{1}{5}} \\
=\sum_{k=0}^{2017} \frac{1}{10}-\frac{1}{2} \sum_{k=0}^{2017} \frac{1}{-5-z_{k}}+\frac{1}{50} \sum_{k=0}^{2017} \frac{1}{-\frac{1}{5}-z_{k}} \\
=\frac{2018}{10}-\frac{P^{\prime}(-5)}{2 P(-5)}+\frac{P^{\prime}\left(-\frac{1}{5}\right)}{50 P\left(-\frac{1}{5}\right)} \\
=\frac{2018 \times 5^{2017}}{5^{2018}-1}.
\end{array}
$$