Difference between revisions of "2020 AMC 12A Problems/Problem 22"

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<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 ==
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== Solution 1 ==
  
 
Square the given equality to yield
 
Square the given equality to yield
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\sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) = \boxed{\frac 7{16}}.
 
\sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) = \boxed{\frac 7{16}}.
 
</cmath>
 
</cmath>
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== Solution 2 (DeMoivre's Formula) ==
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We rewrite
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<cmath>(2+i)^n=\left(\sqrt{5}\left(\frac{2}{\sqrt{5}}+\frac{i}{\sqrt{5}}\right)\right)^n=5^{\frac{n}{2}}e^{in\tan^{-1}\left(\frac{1}{2}\right)}=5^{\frac{n}{2}}\left(\cos \left(n\tan^{-1}\left(\frac{1}{2}\right)\right)+i\sin\left(n\tan^{-1}\left(\frac{1}{2}\right)\right)\right)</cmath>
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by DeMoivre's Formula. Letting <math>\theta=\tan^{-1}\left(\frac{1}{2}\right)</math>, we know that <math>a_n=5^{\frac{n}{2}}\cos \left(n\theta\right)</math> and <math>b_n=5^{\frac{n}{2}}\sin \left(n\theta\right)</math>. The desired sum then turns into
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<cmath>\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^n\cos\left(n\theta\right)\sin\left(n\theta\right)</cmath>
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<cmath>=\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^n\sin\left(2n\theta\right)=\frac{1}{2}\text{Im}\left(\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^ne^{2in\theta}\right)</cmath>
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This is now an infinite geometric series! After finding <math>\sin(2\theta)=2\cdot \frac{2}{\sqrt{5}}\cdot \frac{1}{\sqrt{5}}=\frac{4}{5}</math>, <math>\cos(2\theta)=\sqrt{1-\sin^2(2\theta)}=\frac{3}{5}</math>, we find
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<cmath>S=\frac{1}{2}\text{Im}\left(\frac{1}{1-\frac{5}{7}\cdot e^{2i\theta}}\right)=\frac{1}{2}\text{Im}\left(\frac{1}{1-\frac{3}{7}-\frac{4}{7}i}\right)=\frac{1}{2}\cdot \frac{\frac{4}{7}}{\frac{32}{49}}=\boxed{\textbf{(B) }\frac{7}{16}}</cmath>
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~ktong
  
 
==See Also==
 
==See Also==

Revision as of 02:05, 4 February 2020

Problem

Let $(a_n)$ and $(b_n)$ be the sequences of real numbers such that \[ (2 + i)^n = a_n + b_ni \]for all integers $n\geq 0$, where $i = \sqrt{-1}$. What is\[\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?\] $\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47$

Solution 1

Square the given equality to yield \[(3 + 4i)^n = (a_n + b_ni)^2 = (a_n^2 - b_n^2) + 2a_nb_ni,\] so $a_nb_n = \tfrac12\operatorname{Im}((3+4i)^n)$ and \[\sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) = \boxed{\frac 7{16}}.\]

Solution 2 (DeMoivre's Formula)

We rewrite \[(2+i)^n=\left(\sqrt{5}\left(\frac{2}{\sqrt{5}}+\frac{i}{\sqrt{5}}\right)\right)^n=5^{\frac{n}{2}}e^{in\tan^{-1}\left(\frac{1}{2}\right)}=5^{\frac{n}{2}}\left(\cos \left(n\tan^{-1}\left(\frac{1}{2}\right)\right)+i\sin\left(n\tan^{-1}\left(\frac{1}{2}\right)\right)\right)\] by DeMoivre's Formula. Letting $\theta=\tan^{-1}\left(\frac{1}{2}\right)$, we know that $a_n=5^{\frac{n}{2}}\cos \left(n\theta\right)$ and $b_n=5^{\frac{n}{2}}\sin \left(n\theta\right)$. The desired sum then turns into \[\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^n\cos\left(n\theta\right)\sin\left(n\theta\right)\] \[=\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^n\sin\left(2n\theta\right)=\frac{1}{2}\text{Im}\left(\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^ne^{2in\theta}\right)\] This is now an infinite geometric series! After finding $\sin(2\theta)=2\cdot \frac{2}{\sqrt{5}}\cdot \frac{1}{\sqrt{5}}=\frac{4}{5}$, $\cos(2\theta)=\sqrt{1-\sin^2(2\theta)}=\frac{3}{5}$, we find \[S=\frac{1}{2}\text{Im}\left(\frac{1}{1-\frac{5}{7}\cdot e^{2i\theta}}\right)=\frac{1}{2}\text{Im}\left(\frac{1}{1-\frac{3}{7}-\frac{4}{7}i}\right)=\frac{1}{2}\cdot \frac{\frac{4}{7}}{\frac{32}{49}}=\boxed{\textbf{(B) }\frac{7}{16}}\] ~ktong

See Also

2020 AMC 12A (ProblemsAnswer KeyResources)
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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