Difference between revisions of "1967 AHSME Problems/Problem 39"
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Revision as of 01:00, 23 September 2014
Problem
Given the sets of consecutive integers 
,
,
,
,
, 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 
be the sum of the elements in the nth set. Then equals:
Solution
See also
| 1967 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 38 |
Followed by Problem 40 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
