2. Given the sequence $\left\{a_{n}\right\}(n \geqslant 1)$ satisfies $a_{n+2}=a_{n+1}-a_{n}$, and $a_{2}=1$. If the sum of the first 1999 terms of this sequence is 2000, then the sum of the first 2000 terms equals ( ).
A. 1999
B. 2000
C. 2001
D. 2002
2. C.
$a_{n+3}=a_{n+2}-a_{n+1}=\left(a_{n+1}-a_{n}\right)-a_{n+1}=-a_{n}$, i.e., $a_{n}+a_{n+3}=0$, so $a_{n}+a_{n+1}+a_{n+2}+$ $a_{n+3}+a_{n+4}+a_{n+5}=0$. Also, $1999=333 \times 6+1,2000=333 \times 6+2$, thus the sum of the first 1999 terms is $a_{1}$. Therefore, the sum of the first 2000 terms is $a_{1}+a_{2}=2001$.