Difference between revisions of "2020 AMC 12A Problems/Problem 22"
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== Problem == | == Problem == | ||
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Let <math>(a_n)</math> and <math>(b_n)</math> be the sequences of real numbers such that | Let <math>(a_n)</math> and <math>(b_n)</math> be the sequences of real numbers such that | ||
<cmath>\[ | <cmath>\[ | ||
(2 + i)^n = a_n + b_ni | (2 + i)^n = a_n + b_ni | ||
\]</cmath>for all integers <math>n\geq 0</math>, where <math>i = \sqrt{-1}</math>. What is<cmath>\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?</cmath> | \]</cmath>for all integers <math>n\geq 0</math>, where <math>i = \sqrt{-1}</math>. What is<cmath>\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?</cmath> | ||
+ | |||
<math>\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47</math> | <math>\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47</math> | ||
== Solution 1 == | == Solution 1 == | ||
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Square the given equality to yield | Square the given equality to yield | ||
<cmath> | <cmath> | ||
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== Solution 2 (DeMoivre's Formula) == | == Solution 2 (DeMoivre's Formula) == | ||
− | Note that <math>(2+i) = \sqrt{5} \cdot \left(\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}}i \right)</math>. Let <math>\theta = \arctan (1/2)</math>, then, we know that < | + | Note that <math>(2+i) = \sqrt{5} \cdot \left(\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}}i \right)</math>. Let <math>\theta = \arctan (1/2)</math>, then, we know that <cmath>(2+i) = \sqrt{5} \cdot \left( \cos \theta + i\sin \theta \right),</cmath> so <cmath>(2+i)^n = (\cos (n \theta) + i\sin (n\theta))(\sqrt{5})^n.</cmath> Therefore, |
+ | <cmath>\begin{align*} | ||
+ | \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} \operatorname{Im} \left( \sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} \right). | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Aha! <math>\sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} </math> is a geometric sequence that evaluates to <math>\frac{1}{1-\frac{5}{7}e^{2\theta i}}</math>! Now we can quickly see that <cmath>\sin(2\theta) = 2 \cdot \sin \theta \cdot \cos \theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5},</cmath> <cmath>\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{4}{5}-\frac{1}{5} = \frac{3}{5}.</cmath> Therefore, <cmath>\frac{1}{1-\frac{5}{7}e^{2\theta i}} = \frac{1}{1 - \frac{5}{7}\left( \frac{3}{5} + \frac{4}{5}i\right)} = \frac{7}{8} + \frac{7}{8}i.</cmath> The imaginary part is <math>\frac{7}{8}</math>, so our answer is <math>\frac{1}{2} \cdot \frac{7}{8} = \boxed{\frac{7}{16}} \Rightarrow \textbf{(B)}</math>. | ||
+ | |||
+ | ~AopsUser101 | ||
− | + | == Solution 3 == | |
− | ==Solution 3== | + | Clearly <math>a_n=\tfrac{(2+i)^n+(2-i)^n}{2}, b_n=\tfrac{(2+i)^n-(2-i)^n}{2i}</math>. So we have <math>\sum_{n\ge 0}\tfrac{a_nb_n}{7^n}=\sum_{n\ge 0}\tfrac{((2+i)^n+(2-i)^n))((2+i)^n-(2-i)^n))}{4i(7^n)}</math>. By linearity, we have the latter is equivalent to <math>\tfrac{1}{4i}\sum_{n\ge 0}\tfrac{[(2+i)^n+(2-i)^n][(2+i)^n-(2-i)^n]}{7^n}</math>. Expanding the summand yields |
− | Clearly <math>a_n=\tfrac{(2+i)^n+(2-i)^n}{2}, b_n=\tfrac{(2+i)^n-(2-i)^n}{ | + | <cmath>\begin{align*} |
+ | \tfrac{1}{4i}\sum_{n\ge 0}\tfrac{(3+4i)^n-(3-4i)^n}{7^n}&=\tfrac{1}{4}[\tfrac{1}{1-(\tfrac{3+4i}{7})}-\tfrac{1}{1-(\tfrac{3-4i}{7})}] \\ | ||
+ | &=\tfrac{1}{4i}[\tfrac{7}{7-(3+4i)}-\tfrac{7}{7-(3-4i)}] \\ | ||
+ | &=\tfrac{1}{4}[\tfrac{7}{4-4i}-\tfrac{7}{4+4i}] \\ | ||
+ | &=\tfrac{1}{4i}[\tfrac{7(4+4i)}{32}-\tfrac{7(4-4i)}{32}]=\tfrac{1}{4}\cdot \tfrac{56}{32} \\ | ||
+ | &=\boxed{\tfrac{7}{16}}\textbf{(B)} | ||
+ | \end{align*}</cmath> | ||
-vsamc | -vsamc | ||
− | |||
+ | |||
+ | == Video Solution by Richard Rusczyk == | ||
+ | https://www.youtube.com/watch?v=OdSTfCDOh5A | ||
+ | |||
+ | - AMBRIGGS | ||
+ | |||
+ | == See Also == | ||
{{AMC12 box|year=2020|ab=A|num-b=21|num-a=23}} | {{AMC12 box|year=2020|ab=A|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:07, 10 November 2024
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
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 |
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 |
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