Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AMC 12A Problems/Problem 22"
m
Line 19: Line 19:
  
 
== 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 <math>(2+i) = \sqrt{5} \cdot \left( \cos \theta + i\sin \theta \right)</math>, so <math>(2+i)^n = (\cos (n \theta) + i\sin (n\theta))(\sqrt{5})^n</math>. Therefore, <math>\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} =</math> <math>\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)</math>. 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>. We can quickly see that <math>\sin(2\theta) = 2 \cdot \sin \theta \cdot \cos \theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5}</math>. <math>\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{4}{5}-\frac{1}{5} = \frac{3}{5}</math>. Therefore, <math>\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</math>. 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}}</math>.(Which is answer choice <math>\textbf{(C)}</math>
+
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 <math>(2+i) = \sqrt{5} \cdot \left( \cos \theta + i\sin \theta \right)</math>, so <math>(2+i)^n = (\cos (n \theta) + i\sin (n\theta))(\sqrt{5})^n</math>. Therefore, <math>\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} =</math> <math>\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)</math>. 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>. We can quickly see that <math>\sin(2\theta) = 2 \cdot \sin \theta \cdot \cos \theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5}</math>. <math>\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{4}{5}-\frac{1}{5} = \frac{3}{5}</math>. Therefore, <math>\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</math>. 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}}</math>.(Which is answer choice <math>\textbf{(B)}</math>
  
−
~AopsUser101, minor edit by vsamc stating that the answer choice is C
+
~AopsUser101, minor edit by vsamc stating that the answer choice is B
 
==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}{2}</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))}{4(7^n)}</math>. By linearity, we have the latter is equivalent to <math>\tfrac{1}{4}\sum_{n\ge 0}\tfrac{[(2+i)^n+(2-i)^n][(2+i)^n-(2-i)^n]}{7^n}</math>. Expanding the summand yields <math>\tfrac{1}{4}\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}{4}[\tfrac{7}{7-(3+4i)}+\tfrac{7}{7-(3-4i)}]=\tfrac{1}{4}[\tfrac{7}{4-4i}+\tfrac{7}{4+4i}]=\tfrac{1}{4}[\tfrac{7(4+4i)}{32}+{7(4-4i)}{32}]=\tfrac{1}{4}\cdot \tfrac{56}{32}=\boxed{\tfrac{7}{16}}\textbf{(C)}</math>
+
Clearly <math>a_n=\tfrac{(2+i)^n+(2-i)^n}{2}, b_n=\tfrac{(2+i)^n-(2-i)^n}{2}</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))}{4(7^n)}</math>. By linearity, we have the latter is equivalent to <math>\tfrac{1}{4}\sum_{n\ge 0}\tfrac{[(2+i)^n+(2-i)^n][(2+i)^n-(2-i)^n]}{7^n}</math>. Expanding the summand yields <math>\tfrac{1}{4}\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}{4}[\tfrac{7}{7-(3+4i)}+\tfrac{7}{7-(3-4i)}]=\tfrac{1}{4}[\tfrac{7}{4-4i}+\tfrac{7}{4+4i}]=\tfrac{1}{4}[\tfrac{7(4+4i)}{32}+{7(4-4i)}{32}]=\tfrac{1}{4}\cdot \tfrac{56}{32}=\boxed{\tfrac{7}{16}}\textbf{(B)}</math>
 
-vsamc
 
-vsamc
 
==See Also==
 
==See Also==

Revision as of 14:39, 13 March 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)

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, $\sum_{n=0}^\infty\frac{a_nb_n}{7^n} = \sum_{n=0}^\infty\frac{\cos(n\theta)\sin(n\theta) (5)^n}{7^n} =$ $\frac{1}{2}\sum_{n=0}^\infty \left( \frac{5}{7}\right)^n \sin (2n\theta) = \frac{1}{2} \text{im} \left( \sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} \right)$. Aha $\sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n}$ is a geometric sequence that evaluates to $\frac{1}{1-\frac{5}{7}e^{2\theta i}}$. 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 $\frac{1}{2} \cdot \frac{7}{8} = \boxed{\frac{7}{16}}$.(Which is answer choice $\textbf{(B)}$

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

Solution 3

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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_22&oldid=119332"