Given $x_{1} 、 x_{2}$ are the two roots of the quadratic equation $x^{2}-b x+c=0$ with real coefficients. Find $x_{1}^{n}+x_{2}^{n}\left(n \in \mathbf{N}_{+}\right)$.
By Vieta's formulas, we have
$$
x_{1}+x_{2}=b, x_{1} x_{2}=c \text {. }
$$

Then $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=b^{2}-2 c$. Let $x_{1}^{n}+x_{2}^{n}=S_{n}$. Then $S_{n}=b S_{n-1}-c S_{n-2}$. Therefore,
$$
S_{3}=b S_{2}-c S_{1}=b\left(b^{2}-2 c\right)-c b=b^{3}-3 b c \text {. }
$$

Similarly, $S_{4}=b^{4}-4 b^{2} c+2 c^{2}$,
$$
S_{\mathrm{S}}=b^{5}-5 b^{3} c+5 b c^{2} \text {. }
$$
......
$$
\begin{array}{l}
S_{9}=b^{9}-9 b^{7} c+27 b^{5} c^{2}-30 b^{3} c^{3}+9 b c^{4}, \\
S_{10}=b^{10}-10 b^{8} c+35 b^{6} c^{2}-50 b^{4} c^{3}+25 b^{2} c^{4}-2 c^{5}
\end{array}
$$

To find the general expression for $S_{n}$, we can relate the coefficients of the expanded forms of $S_{9}$ and $S_{10}$ to binomial coefficients. The expanded forms of $S_{9}$ and $S_{10}$ are as follows:
$$
\begin{aligned}
S_{9}= & b^{9}-\mathrm{C}_{9}^{1} b^{7} c+\left(\mathrm{C}_{8}^{2}-\mathrm{C}_{6}^{0}\right) b^{5} c^{2}- \\
& \left(\mathrm{C}_{7}^{3}-\mathrm{C}_{5}^{1}\right) b^{3} c^{3}+\left(\mathrm{C}_{6}^{4}-\mathrm{C}_{4}^{2}\right) b c^{4}, \\
S_{10}= & b^{10}-\mathrm{C}_{10}^{1} b^{8} c+\left(\mathrm{C}_{9}^{2}-\mathrm{C}_{7}^{0}\right) b^{6} c^{2}- \\
& \left(\mathrm{C}_{8}^{3}-\mathrm{C}_{6}^{1}\right) b^{4} c^{3}+\left(\mathrm{C}_{7}^{4}-\mathrm{C}_{5}^{2}\right) b^{2} c^{4}- \\
& \left(\mathrm{C}_{6}^{5}-\mathrm{C}_{4}^{3}\right) c^{5} .
\end{aligned}
$$

Thus, we conjecture:
$$
S_{n}=\sum_{i=0}^{\left[\frac{n}{2}\right]}(-1)^{i}\left(\mathrm{C}_{n+1-i}^{i}-\mathrm{C}_{n-1-i}^{i-2}\right) b^{n-2 i} c^{i},
$$

where $[x]$ is the floor function; and we define:
If $m<0, \mathrm{C}_{n}^{m}=0$.
We will prove this conjecture by mathematical induction.
For $n=1, 2$, it is clearly true.
Assume the conjecture holds for $n=k-1, k$.
For $n=k+1$,
$$
\begin{array}{l} 
S_{k+1}=b S_{k}-c S_{k-1} \\
= \sum_{i=0}^{\left[\frac{k}{2}\right]}(-1)^{i}\left(\mathrm{C}_{k+1-i}^{i}-\mathrm{C}_{k-1-i}^{i-2}\right) b^{k+1-2 i} c^{i} . \\
{\left[\frac{k-1}{2}\right] } \\
(-1)^{i}\left(C_{k-i}^{i}-C_{k-2-i}^{-2}\right) b^{k-1-2 i} c^{i+1} .
\end{array}
$$

When $k$ is even,
$$
\begin{array}{l} 
S_{k+1}=b^{k+1}+\sum_{i=1}^{\frac{k}{2}}(-1)^{i}\left(\mathrm{C}_{k+1-i}^{i}-\mathrm{C}_{k-1-i}^{i-2}\right) b^{k+1-2 i} c^{i}- \\
\sum_{i=0}^{\frac{k}{2}-1}(-1)^{i}\left(\mathrm{C}_{k-i}^{1}-\mathrm{C}_{k-2-i}^{i-2}\right) b^{k-1-2 i} c^{i+} \\
= b^{k+1}+\sum_{i=1}^{\frac{k}{2}}(-1)^{i}\left(\mathrm{C}_{k+1-i}^{i}-\mathrm{C}_{k-1-i}^{i-2}\right) b^{k+1-2 i} c^{i}+ \\
\sum_{i=1}^{\frac{k}{2}}(-1)^{i}\left(\mathrm{C}_{k+1-i}^{i-1}-\mathrm{C}_{k-1-i}^{i-3}\right) b^{k+1-2 i} c^{i} \\
= b^{k+1}+\sum_{i=1}^{\frac{k}{2}}(-1)^{i}\left(\mathrm{C}_{k+2-i}^{i}-\mathrm{C}_{k-i}^{i-2}\right) b^{k+1-2} c^{i} \\
= {\left[\frac{k+1}{2}\right] } \\
=(-1)^{i}\left(\mathrm{C}_{k+2-i}^{i}-\mathrm{C}_{k-i}^{i-2}\right) b^{k+1-2 i} c^{i} .
\end{array}
$$

When $k$ is odd, it can be similarly proven.
Therefore, $x_{1}^{n}+x_{2}^{n}$
$$
=\sum_{i=0}^{\left[\frac{n}{2}\right]}(-1)^{i}\left(\mathrm{C}_{n+1-i}^{i}-\mathrm{C}_{n-1-i}^{i-2}\right) b^{n-2 i} c^{i} .
$$