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

(→‎Solution 3)
m (→‎Solution 2 (DeMoivre's Formula))
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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} \text{im} \left( \sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} \right).
+
&=\frac{1}{2} \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 15:41, 8 November 2022

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)

Note that $(2+i) = \sqrt{5} \cdot \left(\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}}i \right)$. Let $\theta = \arctan (1/2)$, then, we know that \[(2+i) = \sqrt{5} \cdot \left( \cos \theta + i\sin \theta \right),\] so \[(2+i)^n = (\cos (n \theta) + i\sin (n\theta))(\sqrt{5})^n.\] Therefore, \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} \Im \left( \sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} \right). \end{align*}

Aha! $\sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n}$ is a geometric sequence that evaluates to $\frac{1}{1-\frac{5}{7}e^{2\theta i}}$! Now we can quickly see that \[\sin(2\theta) = 2 \cdot \sin \theta \cdot \cos \theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5},\] \[\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{4}{5}-\frac{1}{5} = \frac{3}{5}.\] Therefore, \[\frac{1}{1-\frac{5}{7}e^{2\theta i}} = \frac{1}{1 - \frac{5}{7}\left( \frac{3}{5} + \frac{4}{5}i\right)} = \frac{7}{8} + \frac{7}{8}i.\] The imaginary part is $\frac{7}{8}$, so our answer is $\frac{1}{2} \cdot \frac{7}{8} = \boxed{\frac{7}{16}} \Rightarrow \textbf{(B)}$.

~AopsUser101, minor edit by vsamc stating that the answer choice is B, revamped by OreoChocolate

Solution 3

Clearly $a_n=\tfrac{(2+i)^n+(2-i)^n}{2}, b_n=\tfrac{(2+i)^n-(2-i)^n}{2i}$. So we have $\sum_{n\ge 0}\tfrac{a_nb_n}{7^n}=\sum_{n\ge 0}\tfrac{((2+i)^n+(2-i)^n))((2+i)^n-(2-i)^n))}{4i(7^n)}$. By linearity, we have the latter is equivalent to $\tfrac{1}{4i}\sum_{n\ge 0}\tfrac{[(2+i)^n+(2-i)^n][(2+i)^n-(2-i)^n]}{7^n}$. Expanding the summand yields \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*} -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

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