3. Group the positive integers as follows: $(1),(2,3),(4,5,6),(7,8,9,10), \cdots$, where each group has one more number than the previous group, and the first number of each group is 1 more than the last number of the previous group. Let $S_{n}$ represent the sum of the numbers in the $n$-th group, $S_{21}$ is ( ).
A. 1113
B. 4641
C. 2925
D. 5082
3. B.

From the conditions, we know that the first number of the 21st group is $a_{1}=1+2+3+\cdots+20+1=211$, and the last number is $a_{21}=$ $a_{1}+20=231$, so $S_{21}=\frac{21}{2}[211+231]=4641$.