$\text{A.}$ $2 < 100 a_{100} < \frac{5}{2}$ $\text{B.}$ $\frac{5}{2} < 100 a_{100} < 3$ $\text{C.}$ $3 < 100 a_{100} < \frac{7}{2}$ $\text{D.}$ $\frac{7}{2} < 100 a_{100} < 4$

B

#### 解析：

$\mid \because a_1=1$, 易得 $a_2=\frac{2}{3} \in(0,1)$, 依次类推可得 $a_n \in(0,1)$

$$\therefore \frac{1}{a_{n+1}}-\frac{1}{a_n}=\frac{1}{3-a_n} > \frac{1}{3} \text {, }$$

$\therefore a_n < \frac{3}{n+2},(n \geq 2)$, 即 $a_{100} < \frac{1}{34}, 100 a_{100} < \frac{100}{34} < 3$,

$$\therefore \frac{1}{a_2}-\frac{1}{a_1}=\frac{1}{3}\left(1+\frac{1}{2}\right), \frac{1}{a_3}-\frac{1}{a_2} < \frac{1}{3}\left(1+\frac{1}{3}\right), \frac{1}{a_4}-\frac{1}{a_3} < \frac{1}{3}\left(1+\frac{1}{4}\right), \ldots$$
$$\frac{1}{a_n}-\frac{1}{a_{n-1}} < \frac{1}{3}\left(1+\frac{1}{n}\right),(n \geq 3),$$

$$\therefore \frac{1}{a_{100}}-1 < 33+\frac{1}{3}\left(\frac{1}{2}+\frac{1}{3}+\mathrm{L}+\frac{1}{99}\right) < 33+\frac{1}{3}\left(\frac{1}{2} \times 4+\frac{1}{6} \times 94\right) < 39 \text {, }$$

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