2020 AMC 12A Problems/Problem 22
Contents
Problem
Let and be the sequences of real numbers such that for all integers , where . What is
Solution 1
Square the given equality to yield so and
Solution 2 (DeMoivre's Formula)
Note that . Let , then, we know that so Therefore,
Aha! is a geometric sequence that evaluates to ! Now we can quickly see that Therefore, The imaginary part is , so our answer is .
~AopsUser101
Solution 3
Clearly . So we have . By linearity, we have the latter is equivalent to . Expanding the summand yields -vsamc
Solution 4 - Author : Shiva Kumar Kannan - IN PROGRESS, PLEASE DO NOT MESS
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=OdSTfCDOh5A
- AMBRIGGS
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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All AMC 12 Problems and Solutions |
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