Example 1. (14th All-Russian Mathematics Competition) Let
$$
\frac{1}{1+\frac{1}{1+\frac{1}{1+}}+\frac{1}{1}}=-\frac{m}{n},
$$

where $m$ and $n$ are coprime natural numbers, and the left side of the equation has 1988 fraction lines. Calculate $m^{2}+m n-n^{2}$.
Let the value of the complex fraction with $k$ layers of fraction lines be $\frac{m_{k}}{n_{k}},\left(m_{k}, n_{k}^{*}\right)=1$, then
$$
\frac{m_{\mathrm{k}+1}}{n_{\mathrm{k}+1}}=\frac{1}{1+\frac{m_{\mathrm{k}}}{n_{\mathrm{k}}}}=\frac{n_{\mathrm{k}}}{m_{\triangle}+n_{\mathrm{k}}},
$$

i.e., $m_{\mathrm{k}+1}=n_{\mathrm{k}}, n_{\mathrm{k}+1}=m_{\mathrm{k}}+n_{\mathrm{b}}$.
Notice that $\frac{m_{1}}{n_{1}}=\frac{1}{1}, m_{1}=1, n_{1}=1$, comparing it with the Fibonacci sequence: $f_{1}=1, f_{2}=1, f_{k+1}$ $=f_{k}+f_{k+1}(k \geqslant 1)$, we can see that,
$$
m_{1}=f_{\mathrm{k}}, \quad n_{\mathrm{k}}=f_{\mathrm{k}+1} \text {. }
$$

Thus, $m^{2}+m n-n^{2}$
$$
\begin{array}{l}
=f_{1988}^{2}+f_{1988} f_{1989}-f_{1989}^{2} \\
=f_{1988}^{2}+f_{1988}\left(f_{1988}+f_{1987}\right) \\
-\left(f_{1987}^{2}+2 f_{1988} f_{1987}+f_{1988}^{2}\right) \\
=-\left(f_{1937}^{2}+f_{1987} f_{1988}-f_{1988}\right) \\
=f_{1980}^{2}+f_{1986} f_{1087}-f_{1987}^{2} \\
=\cdots \cdots
\end{array}
$$
$$
\begin{array}{l}
=f_{2}^{2}+f_{2} f_{3}-f_{3}^{2} \\
=1^{2}+1 \cdot 2-2^{2}=-1 .
\end{array}
$$