2023 AMC 12A Problems/Problem 25
Contents
Problem
There is a unique sequence of integers such that whenever is defined. What is
Solution 1
By equating real and imaginary parts:
This problem is the same as problem 7.64 in the Art of Problem Solving textbook Precalculus chapter 7 that asks to prove
Solution 2 (Formula of tanx)
Note that , where k is odd and the sign of each term alternates between positive and negative. To realize this during the test, you should know the formulas of and , and can notice the pattern from that. The expression given essentially matches the formula of exactly. is evidently equivalent to , or 1. However, it could be positive or negative. Notice that in the numerator, whenever the exponent of the tangent term is congruent to 1 mod 4, the term is positive. Whenever the exponent of the tangent term is 3 mod 4, the term is negative. 2023, which is assigned to k, is congruent to 3 mod 4. This means that the term of is .
Notice: If you have time and don't know and , you'd have to keep deriving until you see the pattern.
~lprado
Solution 3
For odd , we have
Thus, for , we have
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
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All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.