2020 AMC 12A Problems/Problem 22
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 . We can quickly see that . . Therefore, . The imaginary part is , so our answer is .(Which is answer choice
~AopsUser101, minor edit by vsamc stating that the answer choice is C
Solution 3
Clearly . So we have . By linearity, we have the latter is equivalent to . Expanding the summand yields =$=\tfrac{1}{4}[\tfrac{7}{7-(3+4i)}+\tfrac{7}{7-(3-4i))}]=\tfrac{1}{4}[\tfrac{7}{4-4i}+\tfrac{7}{4+4i}\=\tfrac{1}{4}[\tfrac{7(4+4i)}{32}+{7(4-4i)}{32}=\tfrac{1}{4}\cdot \tfrac{56}{32}=\boxed{\tfrac{7}{16}}\textbf{(C)}$ (Error compiling LaTeX. Unknown error_msg) -vsamc
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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