Example 10 Given a sequence $\left\{a_{n}\right\}: a_{1}, a_{2}, \cdots$, where $a_{1}$ is a natural number, and $a_{n+1}=\left[1.5 a_{n}\right]+1([x]$ represents the greatest integer not exceeding the real number $x)$. Can we choose the value of $a_{1}$ so that the first $10^{5}$ terms of the sequence $\left\{a_{n}\right\}$ are even, and the 100001st term is odd?
(1981, Federal German Mathematical Competition)
Solution: If $1 \leqslant n \leqslant 10^{5}$, $a_{n}$ is even, then $\left[1.5 a_{n}\right]=1.5 a_{n}$.
Thus, $a_{n+1}=1.5 a_{n}+1\left(1 \leqslant n \leqslant 10^{5}\right)$.
Since the fixed point of $f(x)=1.5 x+1$ is $x_{0}=$ -2, by Theorem 1, we have
$$
a_{n}=\left(\frac{3}{2}\right)^{n-1}\left(a_{1}+2\right)-2\left(1 \leqslant n \leqslant 10^{5}\right) \text {. }
$$

Therefore, by taking $a_{1}=2^{100000} \cdot N-2$ (where $N$ is any odd number), we have $a_{n}=3^{n-1} \times 2^{100001-n} \cdot N-2$ (1 $\left.\leqslant n \leqslant 10^{5}\right)$. At this time, $a_{1}, a_{2}, \cdots, a_{100000}$ are even, and $a_{100001}=3^{100000} \cdot N-2$ is odd.

Theorem 2 Let $f(x)=\frac{A x+B}{C x+D}$ have two fixed points $x_{1} 、 x_{2}$, and the sequence $\left\{a_{n}\right\}$ is defined by $a_{n+1}=f\left(a_{n}\right)$.
(1) If $x_{1} \neq x_{2}$, then the sequence $\left\{\frac{a_{n}-x_{1}}{a_{n}-x_{2}}\right\}$ is a geometric sequence;
(2) If $x_{1}=x_{2}$, then the sequence $\left\{\frac{1}{a_{n}-x_{1}}\right\}$ is an arithmetic sequence.

Proof: Since $x_{k}(k=1,2)$ are fixed points of $f(x)$, i.e., $x_{k}=\frac{A x_{k}+B}{C x_{k}+D}$, which is equivalent to
$$
C x_{k}^{2}+(D-A) x_{k}-B=0 \text {, }
$$

then $D x_{k}-B=\left(A-C x_{k}\right) x_{k}$.
Thus, we have
$$
\begin{array}{l}
\text { (1) } \frac{a_{n+1}-x_{1}}{a_{n+1}-x_{2}}=\frac{\frac{A a_{n}+B}{C a_{n}+D}-x_{1}}{\frac{A a_{n}+B}{C a_{n}+D}-x_{2}} \\
=\frac{A a_{n}+B-x_{1}\left(C a_{n}+D\right)}{A a_{n}+B-x_{2}\left(C a_{n}+D\right)} \\
=\frac{A-C x_{1}}{A-C x_{2}} \cdot \frac{a_{n}-x_{1}}{a_{n}-x_{2}} .
\end{array}
$$

Since $A 、 C 、 x_{1} 、 x_{2}$ are all constants, the sequence $\left\{\frac{a_{n}-x_{1}}{a_{n}-x_{2}}\right\}$ is a geometric sequence.
$$
\begin{array}{l}
\text { (2) } \frac{1}{a_{n+1}-x_{1}}=\frac{1}{\frac{A a_{n}+B}{C a_{n}+D}-x_{1}} \\
=\frac{C a_{n}+D}{A a_{n}+B-x_{1}\left(C a_{n}+D\right)} \\
=\frac{C a_{n}+D}{\left(A-C x_{1}\right) a_{n}+B-D x_{1}} \\
=\frac{C a_{n}+D}{\left(A-C x_{1}\right) a_{n}+\left(C x_{1}-A\right) x_{1}} \\
=\frac{\left(C a_{n}-C x_{1}\right)+\left(C x_{1}+D\right)}{\left(A-C x_{1}\right)\left(a_{n}-x_{1}\right)} \\
=\frac{C}{A-C x_{1}}+\frac{C x_{1}+D}{\left(A-C x_{1}\right)\left(a_{n}-x_{1}\right)} .
\end{array}
$$

Since equation (1) has two equal roots $x_{1}=x_{2}$, by the quadratic formula, we get $x_{1}=\frac{A-D}{2 C}$, substituting into equation (2), we get
$$
\begin{array}{l}
\frac{1}{a_{n+1}-x_{1}}=\frac{C}{A-C x_{1}}+\frac{C x_{1}+D}{\left(A-C x_{1}\right)\left(a_{n}-x_{1}\right)} \\
=\frac{2 C}{A+D}+\frac{1}{a_{n}-x_{1}} .
\end{array}
$$

Therefore, the sequence $\left\{\frac{1}{a_{n}-x_{1}}\right\}$ is an arithmetic sequence.