3. The sequence $a_{0}=134, a_{1}=150, a_{k+1}=a_{k-1}-\frac{k}{a_{k}}(k=1,2, \cdots, n-1)$. If $a_{n}=0$, then $n$ is
A. 20
B. 201
C. 2017
D. 20101
$$
a_{k+1}=\frac{a_{k-1} a_{k}-k}{a_{k}} \Rightarrow a_{k} a_{k+1}-a_{k-1} a_{k}=-k \Rightarrow a_{k} a_{k+1}=a_{0} a_{1}-\frac{k(k+1)}{2} \text {, }
$$

Thus $a_{n}=0 \Rightarrow \frac{k(k+1)}{2}=134 \cdot 150 \Rightarrow k+1=201$, so the answer is $B$.