$15 \cdot 6$ Given the sequence of sets of consecutive integers $\{1\},\{2,3\},\{4,5,6\},\{7,8,9,10\}, \cdots$, where each set contains one more element than the previous one, and the first element of each set is 1 more than the last element of the previous set. Let $S_{n}$ be the sum of the elements in the $n$-th set. Then $S_{21}$ is equal to
(A) 1113 .
(B) 4641 .
(C) 5082 .
(D) 53361 .
(E) None of the above.
(18th American High School Mathematics Examination, 1967)
[Solution] Suppose the $n$-th set system consists of $n$ consecutive integers, and the last one (i.e., 1, 3, 6, 10, ...) is the sum of the first $n$ natural numbers $\frac{n(n+1)}{2}$.
Then the first term of the $n$-th set is $\frac{n(n+1)}{2}-(n-1)$, and the sum of all its elements is
$$
S_{n}=n \cdot \frac{n(n+1)}{2}-\frac{n(n-1)}{2}=\frac{1}{2} n\left(n^{2}+1\right) \text {. }
$$

When $n=21$, $S_{21}=\frac{1}{2} \cdot 21 \cdot\left(21^{2}+1\right)=4641$.
Therefore, the answer is $(B)$.