3. Given the sequence
$$
\begin{array}{l}
a_{0}=134, a_{1}=150, \\
a_{k+1}=a_{k-1}-\frac{k}{a_{k}}(k=1,2, \cdots, n-1) .
\end{array}
$$

If $a_{n}=0$, then $n$ is ( ).
(A) 20
(B) 201
(C) 2017
(D) 20101
3. B.

From $a_{n}=a_{n-2}-\frac{n-1}{a_{n-1}}=0$, we know $a_{n-2} a_{n-1}=n-1$.
Also, from equation (1), we know $a_{k-1} a_{k}-a_{k} a_{k+1}=k$.
Therefore, $a_{0} a_{1}=a_{n-2} a_{n-1}+\sum_{k=1}^{n-2}\left(a_{k-1} a_{k}-a_{k} a_{k+1}\right)$
$$
\begin{array}{l}
=\sum_{k=1}^{n-1} k=\frac{n(n-1)}{2} \\
\Rightarrow n(n-1)=40200 \Rightarrow n=201 .
\end{array}
$$