1. Given $a_{n}=\log _{n}(n+1)$, let $\sum_{n=2}^{1023} \frac{1}{\log _{100} a_{n}}$ $=\frac{q}{p}$, where $p, q$ are integers, and $(p, q)=1$. Then $p+q=(\quad)$.
(A) 3
(B) 1023
(C) 2000
(D) 2001
-1 (A).
$$
\begin{array}{l}
\sum_{n=2}^{1023} \frac{1}{\log _{a_{n}} 100}=\sum_{n=2}^{1023} \log _{100} a_{n} \\
=\log _{100}\left(a_{2} a_{3} \cdots a_{1023}\right) . \\
\text{Since } a_{n}=\log _{n}(n+1)=\frac{\lg (n+1)}{\lg n}, \\
\text{therefore } a_{2} a_{3} \cdots a_{1023}=\frac{\lg 3}{\lg 2} \cdot \frac{\lg 4}{\lg 3} \cdots \cdots \frac{\lg 1024}{\lg 1023}=\frac{\lg 1024}{\lg 2} \\
=\log _{2} 1024=10 . \\
\text{Hence } \sum_{n=2}^{1023} \frac{1}{\log _{a_{n}} 100}=\log _{100} 10=\frac{1}{2}, \text{ i.e., } \frac{q}{p}=\frac{1}{2} . \\
\text{Since } (p, q)=1, \\
\text{therefore } p=2, q=1 .
\end{array}
$$

Thus $p+q=3$.