Difference between revisions of "2020 AMC 12A Problems/Problem 22"
(Created page with "== Problem == Let <math>(a_n)</math> and <math>(b_n)</math> be the sequences of real numbers such that <cmath>\[ (2 + i)^n = a_n + b_ni \]</cmath>for all integers <math>n\geq...") |
(→Solution 4 - Author : Shiva Kumar Kannan - IN PROGRESS, PLEASE DO NOT MESS) |
||
| (30 intermediate revisions by 12 users not shown) | |||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | |||
Let <math>(a_n)</math> and <math>(b_n)</math> be the sequences of real numbers such that | Let <math>(a_n)</math> and <math>(b_n)</math> be the sequences of real numbers such that | ||
<cmath>\[ | <cmath>\[ | ||
(2 + i)^n = a_n + b_ni | (2 + i)^n = a_n + b_ni | ||
\]</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 <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> | ||
| + | |||
<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 | ||
<cmath> | <cmath> | ||
| Line 18: | Line 17: | ||
</cmath> | </cmath> | ||
| − | == | + | == 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, | ||
| + | <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} \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>. | ||
| + | |||
| + | ~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 | ||
| + | <cmath>\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*}</cmath> | ||
| + | -vsamc | ||
| + | |||
| + | |||
| + | == Video Solution by Richard Rusczyk == | ||
| + | https://www.youtube.com/watch?v=OdSTfCDOh5A | ||
| + | |||
| + | - AMBRIGGS | ||
| + | == See Also == | ||
{{AMC12 box|year=2020|ab=A|num-b=21|num-a=23}} | {{AMC12 box|year=2020|ab=A|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 23:07, 10 November 2024
Contents
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)
Note that 
. Let 
, then, we know that ![\[(2+i) = \sqrt{5} \cdot \left( \cos \theta + i\sin \theta \right),\]](http://latex.artofproblemsolving.com/c/a/f/caf67b2c96ae72039aea91f83a88034fb4d656dd.png)
so ![\[(2+i)^n = (\cos (n \theta) + i\sin (n\theta))(\sqrt{5})^n.\]](http://latex.artofproblemsolving.com/f/a/c/fac90b398f88a91f398e5068521866e7bb01fa3b.png)
Therefore,
Aha! 
is a geometric sequence that evaluates to 
! Now we can quickly see that ![\[\sin(2\theta) = 2 \cdot \sin \theta \cdot \cos \theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5},\]](http://latex.artofproblemsolving.com/1/c/f/1cf16f60a47d91d42ced30101e20f751758a474b.png)
![\[\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{4}{5}-\frac{1}{5} = \frac{3}{5}.\]](http://latex.artofproblemsolving.com/9/7/2/972578bc81fdf63400ca33481570b230c81c05dc.png)
Therefore, ![\[\frac{1}{1-\frac{5}{7}e^{2\theta i}} = \frac{1}{1 - \frac{5}{7}\left( \frac{3}{5} + \frac{4}{5}i\right)} = \frac{7}{8} + \frac{7}{8}i.\]](http://latex.artofproblemsolving.com/9/4/5/9451d5dd164ea7e568ecd31cda21a0acc4edff07.png)
The imaginary part is 
, so our answer is 
.
~AopsUser101
Solution 3
Clearly 
. So we have 
. By linearity, we have the latter is equivalent to ![$\tfrac{1}{4i}\sum_{n\ge 0}\tfrac{[(2+i)^n+(2-i)^n][(2+i)^n-(2-i)^n]}{7^n}$](http://latex.artofproblemsolving.com/b/6/f/b6f927c5a1a927f7cddce65558ea83db98a45809.png)
. 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*}](http://latex.artofproblemsolving.com/3/3/0/330acdd1524e0fed5a5b28fdcb1c441b01c04bd9.png)
-vsamc
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=OdSTfCDOh5A
- 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
