1969 AHSME Problems/Problem 16
Problem
When , is expanded by the binomial theorem, it is found that when , where is a positive integer, the sum of the second and third terms is zero. Then equals:
Solution
Since , we can write as . Expanding, the second term is , and the third term is , so we can write the equation Simplifying and multiplying by two to remove the denominator, we get Factoring, we get Dividing by gives Since it is given that , cannot equal 0, so we can divide by n, which gives Solving for gives so the answer is .
See also
1969 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.