Difference between revisions of "2020 AMC 12A Problems/Problem 22"
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\sum_{n=0}^\infty\frac{a_nb_n}{7^n} &= \sum_{n=0}^\infty\frac{\cos(n\theta)\sin(n\theta) (5)^n}{7^n} \ | \sum_{n=0}^\infty\frac{a_nb_n}{7^n} &= \sum_{n=0}^\infty\frac{\cos(n\theta)\sin(n\theta) (5)^n}{7^n} \ | ||
&=\frac{1}{2}\sum_{n=0}^\infty \left( \frac{5}{7}\right)^n \sin (2n\theta)\ | &=\frac{1}{2}\sum_{n=0}^\infty \left( \frac{5}{7}\right)^n \sin (2n\theta)\ | ||
− | &=\frac{1}{2} \Im \left( \sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} \right). | + | &=\frac{1}{2} \operatorname{Im} \left( \sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} \right). |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Revision as of 15:46, 8 November 2022
Contents
[hide]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, minor edit by vsamc stating that the answer choice is B, revamped by OreoChocolate
Solution 3
Clearly . So we have . By linearity, we have the latter is equivalent to . Expanding the summand yields -vsamc
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=OdSTfCDOh5A&list=PLyhPcpM8aMvJvwA2kypmfdtlxH90ShZCc&index=2 - AMBRIGGS
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.