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 == | + | == Solution 1 == |
Square the given equality to yield | Square the given equality to yield | ||
| Line 17: | Line 17: | ||
\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> | ||
| + | |||
| + | == Solution 2 (DeMoivre's Formula) == | ||
| + | We rewrite | ||
| + | <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> | ||
| + | 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 | ||
| + | <cmath>\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^n\cos\left(n\theta\right)\sin\left(n\theta\right)</cmath> | ||
| + | <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> | ||
| + | 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 | ||
| + | <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> | ||
| + | ~ktong | ||
==See Also== | ==See Also== | ||
Revision as of 02:05, 4 February 2020
Problem
Let 
and 
be the sequences of real numbers such that
![\[ (2 + i)^n = a_n + b_ni \]](http://latex.artofproblemsolving.com/9/9/1/9917b4f9dba198e73634715eaa0ff52b7e9e3ba5.png)
for all integers 
, where 
. What is![\[\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?\]](http://latex.artofproblemsolving.com/a/3/a/a3a51079920de0ca06d385c58399f194a5ff90cb.png)
Solution 1
Square the given equality to yield
![\[(3 + 4i)^n = (a_n + b_ni)^2 = (a_n^2 - b_n^2) + 2a_nb_ni,\]](http://latex.artofproblemsolving.com/e/a/f/eafed2044252dfe02ba40174e3fdf1321f502761.png)
so 
and
![\[\sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) = \boxed{\frac 7{16}}.\]](http://latex.artofproblemsolving.com/7/f/8/7f85a857c032580b71599a24b712617c3a49855c.png)
Solution 2 (DeMoivre's Formula)
We rewrite
![\[(2+i)^n=\left(\sqrt{5}\left(\frac{2}{\sqrt{5}}+\frac{i}{\sqrt{5}}\right)\right)^n=5^{\frac{n}{2}}e^{in\tan^{-1}\left(\frac{1}{2}\right)}=5^{\frac{n}{2}}\left(\cos \left(n\tan^{-1}\left(\frac{1}{2}\right)\right)+i\sin\left(n\tan^{-1}\left(\frac{1}{2}\right)\right)\right)\]](http://latex.artofproblemsolving.com/0/0/6/00639856b9a02d24bacf06260334d3133ba59fac.png)
by DeMoivre's Formula. Letting 
, we know that 
and 
. The desired sum then turns into
![\[\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^n\cos\left(n\theta\right)\sin\left(n\theta\right)\]](http://latex.artofproblemsolving.com/7/2/0/7206ade687a73ebbf63c0038901102da873cf560.png)
![\[=\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^n\sin\left(2n\theta\right)=\frac{1}{2}\text{Im}\left(\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^ne^{2in\theta}\right)\]](http://latex.artofproblemsolving.com/3/1/b/31b23b557730f5472c1790635e375f02e2cde5fa.png)
This is now an infinite geometric series! After finding 
, 
, we find
![\[S=\frac{1}{2}\text{Im}\left(\frac{1}{1-\frac{5}{7}\cdot e^{2i\theta}}\right)=\frac{1}{2}\text{Im}\left(\frac{1}{1-\frac{3}{7}-\frac{4}{7}i}\right)=\frac{1}{2}\cdot \frac{\frac{4}{7}}{\frac{32}{49}}=\boxed{\textbf{(B) }\frac{7}{16}}\]](http://latex.artofproblemsolving.com/9/0/b/90b33f81e9174bb4f449dc7abb8cc5c6b80667e3.png)
~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. 
