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

2020 AMC 12A Problems/Problem 22

Revision as of 01:11, 4 February 2020 by Ktong (talk | contribs) (→‎Solution 2 (DeMoivre's Formula))

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 \[\sum_{n=0}^{\infty}\frac{a_nb_n}{7^n}=\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^n\cos\left(n\theta\right)\sin\left(n\theta\right)\] \[=\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^n\sin\left(2n\theta\right)=\frac{1}{2}\text{Im}\left(\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^ne^{2in\theta}\right)\] 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 (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=116834"