1967 AHSME Problems/Problem 39

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Problem

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:

$\textbf{(A)}\ 1113\qquad \textbf{(B)}\ 4641 \qquad \textbf{(C)}\ 5082\qquad \textbf{(D)}\ 53361\qquad \textbf{(E)}\ \text{none of these}$

Solution

The last element of the 21st set is $\frac{21\times 22}{2}=231$, and hence the first element of the 21st set is $211$. So $S_{21}=\frac{231\times 232}{2}-\frac{210\times 211}{2}=4641$, hence our answer is $\fbox{B}$

See also

1967 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 38
Followed by
Problem 40
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