已知数列 $\left\{a_{n}\right\}, a_{1}=2, a_{n+1}=1-\frac{1}{a_{n}}$, 则 $a_{1} a_{2} \cdots a_{99}=(\quad)$
答案: C

解析: 因为 $a_{1}=2, a_{n+1}=1-\frac{1}{a_{n}}, a_{2}=\frac{1}{2}, a_{3}=-1, a_{4}=2, \cdots$. 故数列 $\left\{a_{n}\right\}$ 周期为 3 , 且 $a_{1} a_{2} a_{3}=-1$, 所以 $a_{1} a_{2} \cdots a_{99}=\left(a_{1} a_{2} a_{3}\right)^{33}=-1$. 故选 C.