Let $n$ be a natural number, $\alpha_{n} \backslash \beta_{n}\left(\alpha_{n}>\beta_{n}\right)$ are the integer parts of the roots of the quadratic equation $x^{2}-2(n+2) x+3(n+1)=0$. Find the value of $\frac{\alpha_{1}}{\beta_{1}}+\frac{\alpha_{2}}{\beta_{2}}+\cdots+\frac{\alpha_{99}}{\beta_{99}}$.
Solution: Let the two roots of the original equation be $x_{n}, x_{n}^{\prime}$, then by the root formula we get
$$
\begin{array}{l}
x_{n}=n+2+\sqrt{n^{2}+n+1}, \\
x_{n}^{\prime}=n+2-\sqrt{n^{2}+n+1} . \\
\because n<\sqrt{n^{2}+n+1}<n+1, \\
2 n+2<n+2+\sqrt{n^{2}} \overline{n+1}<2 n+3,
\end{array}
$$
E. $\quad<n+2-\sqrt{n^{2}+n+1}<2$.
Therefore, the integer parts of $x_{n}$ and $x_{n}^{\prime}$ are $2 n+2$ and $1$ respectively.
Hence $\alpha_{n}=2 n+2, \beta_{n}=1$.
$$
\begin{array}{l}
\therefore \frac{\alpha_{1}}{\beta_{1}}+\frac{\alpha_{2}}{\beta_{2}}+\cdots+\frac{\alpha_{99}}{\beta_{99}} \\
=2(2+3+\cdots+100) \\
=10098 .
\end{array}
$$