2. The cyclic sequence $\left\{u_{n}\right\}, u_{1}=a$ ( $a$ is any positive number) , $u_{n+1}=-\frac{1}{u_{n}+1}, n=1,2,3, \cdots$, among the following values, the value of $n$ that can make $u_{n}=a$ is
A. 14
B. 15
C. 16
D. 17
2. C Given that $u_{1}=a, u_{2}=-\frac{1}{a+1}$,
$$
u_{3}=-\frac{1}{1-\frac{1}{a+1}}=-1-\frac{1}{a}, u_{4}=-\frac{1}{1-1-\frac{1}{a}}=a \text {. }
$$

Therefore, the period of $\left\{u_{n}\right\}$ is 3, and since $3 \mid(16-1)$, the answer is C.