Competition problem — Number theory

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Number theory Difficulty 2.2 Junior Find the answer

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:

Pick one

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 $\fbox{B}$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.