AHSME 1967 Problem 39

Given the sets of consecutive integers $\{1\}$,$\{2, 3\}$,$\{4,5,6\}$,$\{7,8,9,10\}$,$\cdots \;$, where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let $S_n$ be the sum of the elements in the nth set. Then $S_{21}$ equals: $\textbf{(A)}\ 1113 \textbf{(B)}\ 4641 \textbf{(C)}\ 5082 \textbf{(D)}\ 53361 \textbf{(E)}\ \text{none of these}$