2020 AMC 12A Problems/Problem 22: Difference between revisions

Latest revision as of 01:34, 8 November 2025

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, $\begin{align*} \sum_{n=0}^\infty\frac{a_nb_n}{7^n} &= \sum_{n=0}^\infty\frac{\cos(n\theta)\sin(n\theta) (5)^n}{7^n} \\ &=\frac{1}{2}\sum_{n=0}^\infty \left( \frac{5}{7}\right)^n \sin (2n\theta)\\ &=\frac{1}{2} \operatorname{Im} \left( \sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} \right). \end{align*}$

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

Solution 3

Clearly $a_n=\frac{(2+i)^n+(2-i)^n}{2}, b_n=\frac{(2+i)^n-(2-i)^n}{2i}$ . So we have $\sum_{n\ge 0}\frac{a_nb_n}{7^n}=\sum_{n\ge 0}\frac{((2+i)^n+(2-i)^n))((2+i)^n-(2-i)^n))}{4i(7^n)}$ . By linearity, we have the latter is equivalent to $\frac{1}{4i}\sum_{n\ge 0}\frac{[(2+i)^n+(2-i)^n][(2+i)^n-(2-i)^n]}{7^n}$ . Expanding the summand yields \begin{align*} \frac{1}{4i}\sum_{n\ge 0}\frac{(3+4i)^n-(3-4i)^n}{7^n}&=\frac{1}{4}[\frac{1}{1-(\frac{3+4i}{7})}-\frac{1}{1-(\frac{3-4i}{7})}] \\ &=\frac{1}{4i}[\frac{7}{7-(3+4i)}-\frac{7}{7-(3-4i)}] \\ &=\frac{1}{4}[\frac{7}{4-4i}-\frac{7}{4+4i}] \\ &=\frac{1}{4i}[\frac{7(4+4i)}{32}-\frac{7(4-4i)}{32}]=\frac{1}{4}\cdot \frac{56}{32} \\ &=\boxed{\frac{7}{16}}\textbf{(B)} \end{align*} -vsamc

Video Solution 1 by Richard Rusczyk

https://www.youtube.com/watch?v=OdSTfCDOh5A

Video Solution 2 by StressedPineapple

https://youtube.com/watch?v=NWBPm3lThH4&t=952s

@@ Line 1: / Line 1: @@
 == Problem ==
-Let <math>(a_n)</math> and <math>(b_n)</math> be the sequences of real numbers such that
+Let <imath>(a_n)</imath> and <imath>(b_n)</imath> be the sequences of real numbers such that
 <cmath>\[
 (2 + i)^n = a_n + b_ni
-\]</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 <imath>n\geq 0</imath>, where <imath>i = \sqrt{-1}</imath>. 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>
+<imath>\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47</imath>
 == Solution 1 ==
@@ Line 12: / Line 12: @@
 (3 + 4i)^n = (a_n + b_ni)^2 = (a_n^2 - b_n^2) + 2a_nb_ni,
 </cmath>
-so <math>a_nb_n = \tfrac12\operatorname{Im}((3+4i)^n)</math> and
+so <imath>a_nb_n = \tfrac12\operatorname{Im}((3+4i)^n)</imath> and
 <cmath>
 \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}}.
@@ Line 18: / Line 18: @@
 == 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 <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,
+Note that <imath>(2+i) = \sqrt{5} \cdot \left(\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}}i \right)</imath>. Let <imath>\theta = \arctan (1/2)</imath>, 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} \text{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>
-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>.
+Aha! <imath>\sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} </imath> is a geometric sequence that evaluates to <imath>\frac{1}{1-\frac{5}{7}e^{2\theta i}}</imath>! 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 <imath>\frac{7}{8}</imath>, so our answer is <imath>\frac{1}{2} \cdot \frac{7}{8} = \boxed{\frac{7}{16}} \Rightarrow \textbf{(B)}</imath>.
-~AopsUser101, minor edit by vsamc stating that the answer choice is B, revamped by OreoChocolate
+~AopsUser101
 == 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 <imath>a_n=\frac{(2+i)^n+(2-i)^n}{2}, b_n=\frac{(2+i)^n-(2-i)^n}{2i}</imath>. So we have <imath>\sum_{n\ge 0}\frac{a_nb_n}{7^n}=\sum_{n\ge 0}\frac{((2+i)^n+(2-i)^n))((2+i)^n-(2-i)^n))}{4i(7^n)}</imath>. By linearity, we have the latter is equivalent to <imath>\frac{1}{4i}\sum_{n\ge 0}\frac{[(2+i)^n+(2-i)^n][(2+i)^n-(2-i)^n]}{7^n}</imath>. Expanding the summand yields
-<cmath>\begin{align*}
+\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})}] \\
+\frac{1}{4i}\sum_{n\ge 0}\frac{(3+4i)^n-(3-4i)^n}{7^n}&=\frac{1}{4}[\frac{1}{1-(\frac{3+4i}{7})}-\frac{1}{1-(\frac{3-4i}{7})}] \\
-&=\tfrac{1}{4i}[\tfrac{7}{7-(3+4i)}-\tfrac{7}{7-(3-4i)}] \\
+&=\frac{1}{4i}[\frac{7}{7-(3+4i)}-\frac{7}{7-(3-4i)}] \\
-&=\tfrac{1}{4}[\tfrac{7}{4-4i}-\tfrac{7}{4+4i}] \\
+&=\frac{1}{4}[\frac{7}{4-4i}-\frac{7}{4+4i}] \\
-&=\tfrac{1}{4i}[\tfrac{7(4+4i)}{32}-\tfrac{7(4-4i)}{32}]=\tfrac{1}{4}\cdot \tfrac{56}{32} \\
+&=\frac{1}{4i}[\frac{7(4+4i)}{32}-\frac{7(4-4i)}{32}]=\frac{1}{4}\cdot \frac{56}{32} \\
-&=\boxed{\tfrac{7}{16}}\textbf{(B)}
+&=\boxed{\frac{7}{16}}\textbf{(B)}
-\end{align*}</cmath>
+\end{align*}
 -vsamc
-== Video Solution by Richard Rusczyk ==
+== Video Solution 1 by Richard Rusczyk ==
-https://www.youtube.com/watch?v=OdSTfCDOh5A&list=PLyhPcpM8aMvJvwA2kypmfdtlxH90ShZCc&index=2
+https://www.youtube.com/watch?v=OdSTfCDOh5A
-- AMBRIGGS
+==Video Solution 2 by StressedPineapple==
+https://youtube.com/watch?v=NWBPm3lThH4&t=952s
 == See Also ==
 {{AMC12 box|year=2020|ab=A|num-b=21|num-a=23}}
 {{MAA Notice}}

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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