66 The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, \sqrt{\frac{1}{a_{n}^{2}}+4}=\frac{1}{a_{n+1}}$, and let $S_{n}=\sum_{i=1}^{n} a_{i}^{2}$. If $S_{2 n+1}-S_{n} \leqslant \frac{t}{30}$ holds for any $n \in \mathbf{N}^{*}$, then the minimum value of the positive integer $t$ is
(A) 10
(B) 9
(C) 8
(D) 7
6) A Hint: Given $\frac{1}{a_{n+1}^{2}}-\frac{1}{a_{n}^{2}}=4$, we can find $a_{n}^{2}=\frac{1}{4 n-3}$. Let $g(n)=S_{2 n+1}-S_{n}$, then
$$
\begin{aligned}
g(n+1)-g(n) & =a_{2 n+2}^{2}+a_{2 n+3}^{2}-a_{n+1}^{2} \\
& =\frac{1}{8 n+5}+\frac{1}{8 n+9}-\frac{1}{4 n+1} \\
& <0,
\end{aligned}
$$

i.e., $g(n)$ is a decreasing function, so
$$
S_{2 n+1}-S_{n} \leqslant g(1)=\frac{14}{45} \leqslant \frac{t}{30},
$$

thus $t \geqslant \frac{28}{3}$, and since $n \in \mathbf{N}^{*}$, the minimum value of $t$ is 10.