Problem: Given the sets of consecutive integers {1},{2,3},{4,5,6},{7,8,9,10},⋯, where each set contains one more element than the preceding one, and where the first element of each succeeding set is one more than the last element of the preceding set. Let Sn​ be the sum of the elements in the nth set. Then S21​ equals:
Answer Choices:
A. 1113
B. 4641
C. 5082
D. 53361
E. none of these
Solution:
The first element in the nth set is 1 more than the sum of the number of elements in the preceding n−1 sets, that is, {1+2+…+n−1}+1=2(n−1)(n)​+1=2n2−n+2​. Since the nth set contains n elements its last element is 2n2−n+2​+n−1=2n2+n​. Therefore, Sn​=2n​(2n2−n+2​+2n2+n​) =2n​(n2+1)∴S21​=221​(212+1)=4641.