Question 114, Given the equation $\mathrm{x}^{10}+(13 \mathrm{x}-1)^{10}=0$ has 5 pairs of conjugate complex roots $\mathrm{r}_{\mathrm{k}} 、 \overline{\mathrm{r}_{\mathrm{k}}}(1 \leq \mathrm{k} \leq 5)$, try to find the value of $\sum_{\mathrm{k}=1}^{5} \frac{1}{\mathrm{r}_{\mathrm{k}} \cdot \mathrm{r}_{\mathrm{k}}}$.
Question 114, Solution: According to the conditions, we know: $\frac{1}{r_{k}} 、 \frac{1}{\mathrm{rk}_{\mathrm{k}}}(1 \leq \mathrm{k} \leq 5)$ are the 5 pairs of conjugate complex roots of the equation
$$
1+(13-y)^{10}=0
$$

Thus, according to Vieta's theorem, we have:
$$
\sum_{\mathrm{k}=1}^{5} \frac{1}{\mathrm{r}_{\mathrm{k}}}+\sum_{\mathrm{k}=1}^{5} \frac{1}{\overline{r_{k}}}=\mathrm{C}_{10}^{1} \times 13=130
$$

Also, because
$$
\begin{array}{l}
1+\left(13-\frac{1}{r^{2}}\right)^{10}=0 \Rightarrow\left|13-\frac{1}{r_{k}}\right|=1 \Rightarrow\left(13-\frac{1}{r_{k}}\right)\left(13-\frac{1}{r_{k}}\right)=1 \\
\Rightarrow \frac{1}{r_{k} \cdot r_{k}}=13\left(\frac{1}{r_{k}}+\frac{1}{r^{k}}\right)-168 \\
\text { Therefore, we know: } \\
\sum_{k=1}^{5} \frac{1}{r_{k} \cdot r_{k}}=\sum_{k=1}^{5}\left[13\left(\frac{1}{r_{k}}+\frac{3}{r_{k}}\right)-168\right]=13 \times 130-168 \times 5=850
\end{array}
$$

Therefore, we know:

In summary, $\sum_{\mathrm{k}=1}^{5} \frac{1}{\mathrm{~N} \cdot 2}=850$
Note: Generally, if $n$ is a positive number, the equation $x^{2 n}+(a x+b)^{2 n}=0 （ a 、 b \in R, a b \neq 0 ）$ has $2 n$ complex roots $r_{k} 、 \overline{r_{k}}(1 \leq k \leq n)$, then $\sum_{k=1}^{n} \frac{1}{n \sqrt{2}}=\frac{n\left(a^{2}+1\right)}{b^{2}}$. For the problem, according to Vieta's theorem, we know:
$$
\sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{r}_{\mathrm{k}}}+\sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{r}_{\mathrm{k}}}=-\frac{2 \mathrm{na}}{\mathrm{b}}
$$

Also, because
$$
\begin{array}{l}
1+\left(a+\frac{b}{r_{k}}\right)^{2 n}=0 \Rightarrow\left|a+\frac{b}{r_{k}}\right|=1 \Rightarrow\left(a+\frac{b}{r_{k}}\right)\left(a+\frac{b}{r_{k}}\right)=1 \\
\Rightarrow \frac{1}{r_{k} \cdot r_{k}}=\frac{1-a^{2}}{b^{2}}-\frac{a}{b}\left(\frac{1}{r_{k}}+\frac{1}{r_{k}}\right)
\end{array}
$$

Therefore, we know:
$$
\sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{r}_{\mathrm{k}} \cdot \overline{r_{k}}}=\sum_{\mathrm{k}=1}^{\mathrm{n}}\left[\frac{1-\mathrm{a}^{2}}{\mathrm{~b}^{2}}-\frac{\mathrm{a}}{\mathrm{b}}\left(\frac{1}{\mathrm{r}_{\mathrm{k}}}+\frac{1}{\overline{r_{k}}}\right)\right]=\frac{\mathrm{n}\left(1-\mathrm{a}^{2}\right)}{\mathrm{b}^{2}}-\frac{\mathrm{a}}{\mathrm{b}} \cdot\left(-\frac{2 \mathrm{na}}{\mathrm{b}}\right)=\frac{\mathrm{n}\left(\mathrm{a}^{2}+1\right)}{\mathrm{b}^{2}}
$$

The proposition is proved!