1. (2005 Hunan Province College Entrance Examination Question) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=0, a_{n-1}=\frac{a_{n}-\sqrt{3}}{\sqrt{3} a_{n}+1}\left(n \in \mathbf{N}^{*}\right)$, then $a_{20}$ equals
A. 0
B. $-\sqrt{3}$
C. $\sqrt{3}$
D. $\frac{\sqrt{3}}{2}$
1. B Solution 1 (Incomplete Inductive Guessing Method) From $a_{1}=0, a_{n+1}=\frac{a_{n}-\sqrt{3}}{\sqrt{3} a_{n}+1}$, we can calculate sequentially, getting $a_{2}=-\sqrt{3}, a_{3}=\sqrt{3}$. $a_{4}=0, \cdots$, thus we know that 3 is a period of the sequence $\left\{a_{n}\right\}$.
Therefore, $a_{2 n}=a_{3 \times 6+2}=a_{2}=-\sqrt{3}$.
Solution 2 (Trigonometric Substitution Iteration Method) From the recursive formula $a_{n+1}=\frac{a_{n}-\sqrt{3}}{1+\sqrt{3} a_{n}}$, considering its structural characteristics, we can think of the trigonometric formula
$$
\tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta} .
$$

Noting that $\tan 60^{\circ}=\sqrt{3}$, we can make a two-angle substitution, letting $a_{n}=\tan x_{n}$.
$$
a_{n+1}=\frac{\tan x_{n}-\tan 60^{\circ}}{1+\tan x_{n} \tan 60^{\circ}}=\tan \left(x_{n}-60^{\circ}\right) .
$$

Similarly, we have
$$
a_{n-2}=\tan \left(x_{n}-60^{\circ}-60^{\circ}\right) .
$$

Proceeding in this manner, we get
$$
a_{n+3}=\tan \left(x_{n}-60^{\circ}-60^{\circ}-60^{\circ}\right)=\tan \left(x_{n}-180^{\circ}\right)=\tan x_{n}=a_{n},
$$

so 3 is a period of the sequence $\left\{a_{n}\right\}$.
Therefore, $a_{20}=a_{3 \times 6-2}=a_{2}=-\sqrt{3}$.
Commentary Solution 2 uses analogy, recalling trigonometric formulas, and clever substitution to find the period, achieving a clever solution to a difficult problem.