Question 17 Given a sequence $a_{1}, a_{2}, a_{3}, \cdots$, where $a_{1}$ is a natural number, and $a_{n+1}=\left[1.5 a_{n}\right]+1$ holds for all natural numbers $n$. Is it possible to choose $a_{1}$ such that the first 100000 terms of this sequence are all even, and the 100001st term is odd?
(Federal Republic of Germany 1981 Mathematics Competition)
We can determine $a_{1}$ such that the first 100000 terms of this sequence are all even, and the 100001st term is odd. For example, we can choose
$$
\begin{array}{l}
a_{1}=2^{100000}-2, \\
a_{2}=\left[\frac{3}{2} a_{1}\right]+1=3 \times 2^{99999}-3+1=3 \times 2^{99999}-2, \\
a_{3}=\left[\frac{3}{2} a_{2}\right]+1=3^{2} \times 2^{99998}-3+1=3^{2} \times 2^{99998}-2, \\
\ldots \ldots \\
a_{99999}=\left[\frac{3}{2} a_{99998}\right]+1=3^{99998} \times 2^{2}-3+1=3^{99998} \times 2^{2}-2, \\
a_{100000}=\left[\frac{3}{2} a_{99999}\right]+1=3^{99999} \times 2-3+1, \\
a_{100001}=\left[\frac{3}{2} a_{100000}\right]+1=3^{100000}-3+1=3^{100000}-2.
\end{array}
$$

Therefore, $a_{1}, a_{2}, \cdots, a_{99999}, a_{100000}$ are all even, and $a_{100001}$ is odd.