60 Let the point representing the complex number $2 k^{2}+i$ on the complex plane be $z_{k}(k=1,2,3, \cdots, 100)$. Then $\arg z_{1}+$ $\arg z_{2}+\cdots+\arg z_{100}=$
A. $\frac{\pi}{4}$
B. $\frac{\pi}{4}-\operatorname{arccot} 201$
C. $\arctan 201$
D. $\frac{\pi}{2}$
60 B. Since
$$
z_{k}=2 k^{2}+i,(k=1,2,3, \cdots, 100),
$$

it follows that $z_{k}$ are all in the first quadrant. Therefore,
$$
\arg z_{k}=\arctan \frac{1}{2 k^{2}}.
$$

Also, since
$$
\frac{1}{2 k-1}>\frac{1}{2 k+1}>0,
$$

we have
$$
\begin{aligned}
& \tan \left(\arctan \frac{1}{2 k-1}-\arctan \frac{1}{2 k+1}\right)=\frac{\frac{1}{2 k-1}-\frac{1}{2 k+1}}{1+\frac{1}{2 k-1} \cdot \frac{1}{2 k+1}} \\
= & \frac{2}{(2 k-1)(2 k+1)+1}=\frac{2}{(2 k)^{2}}=\frac{1}{2 k^{2}},
\end{aligned}
$$

which means
$$
\arctan \frac{1}{2 k^{2}}=\arctan \frac{1}{2 k-1}-\arctan \frac{1}{2 k+1}.
$$

Therefore,
$$
\begin{aligned}
\operatorname{arc} z_{1}+\operatorname{arc} z_{2}+\cdots+\operatorname{arc} z_{100} & =\sum_{k=1}^{100}\left(\arctan \frac{1}{2 k-1}-\arctan \frac{1}{2 k+1}\right) \\
& =\arctan \frac{1}{2 \times 1-1}-\arctan \frac{1}{2 \times 100+1} \\
& =\frac{\pi}{4}-\operatorname{arccot} 201.
\end{aligned}
$$