4. Arrange the numbers $\{2 n+1\}$ in a cycle of: one, two, three, four terms: (3), $(5,7),(9,11,13)$, $(15,17,19,21),(23),(25,27),(29,31$, $33),(35,37,39,41),(43), \cdots$, then the sum of the numbers in the 100th parenthesis is $\qquad$.
4. The 100th parenthesis

is the fine one in the
25th group,
the subsequence of parentheses in.
Accordingly, it forms
an arithmetic sequence
with the first term being 72 and the
common difference being 80. Therefore, its sum is $72+$ (25-
$$
\text { 1) } \times 80=1992 \text {. }
$$