4. If a sequence of numbers $a_{1}, a_{2}, \cdots$ satisfies that for any positive integer $n$ there is $a_{1}+a_{2}+\cdots+a_{n}=n^{3}$, then the value of $\frac{1}{a_{2}-1}+\frac{1}{a_{3}-1}+\cdots+\frac{1}{a_{100}-1}$ is ( ).
(A) $\frac{33}{100}$
(B) $\frac{11}{100}$
(C) $\frac{11}{99}$
(D) $\frac{33}{101}$
4. A.

When $n \geqslant 2$, we have
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n}=n^{3}, \\
a_{1}+a_{2}+\cdots+a_{n-1}=(n-1)^{3} .
\end{array}
$$

Subtracting the two equations gives $a_{n}=3 n^{2}-3 n+1$.
$$
\begin{array}{l}
\text { Then } \frac{1}{a_{n}-1}=\frac{1}{3 n(n-1)} \\
=\frac{1}{3}\left(\frac{1}{n-1}-\frac{1}{n}\right)(n=2,3, \cdots) . \\
\text { Therefore, } \frac{1}{a_{2}-1}+\frac{1}{a_{3}-1}+\cdots+\frac{1}{a_{100}-1} \\
=\frac{1}{3}\left(1-\frac{1}{2}\right)+\frac{1}{3}\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\frac{1}{3}\left(\frac{1}{99}-\frac{1}{100}\right) \\
=\frac{33}{100} .
\end{array}
$$