1. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=2, a_{n+1}=-\frac{1}{a_{n}+1}$, then $a_{2001}$ equals
A. $-\frac{3}{2}$
B. $-\frac{1}{3}$
C. 1
D. 2
1. Choose A.

Slightly explained: From $a_{n+1}=-\frac{1}{a_{n}+1}$, we get $a_{n+2}=-\frac{1}{a_{n+1}+1}=-\frac{1}{-\frac{1}{a_{n}+1}+1}=-1-\frac{1}{a_{n}}, a_{n+3}=$ $-\frac{1}{a_{n+2}+1}=-\frac{1}{-1-\frac{1}{a_{n}}+1}=a_{n}$. Then $a_{2001}=a_{666 \cdot 3+3}=a_{3}=-\frac{1}{a_{2}+1}=-\frac{1}{-\frac{1}{a_{1}+1}+1}$ $=-\frac{3}{2}$.