Difference between revisions of "2020 AMC 12A Problems/Problem 22"
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== Problem == | == Problem == | ||
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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 1 == | == Solution 1 == | ||
− | |||
Square the given equality to yield | Square the given equality to yield | ||
<cmath> | <cmath> | ||
Line 23: | Line 22: | ||
~AopsUser101, minor edit by vsamc stating that the answer choice is B, revamped by OreoChocolate | ~AopsUser101, minor edit by vsamc stating that the answer choice is B, revamped by OreoChocolate | ||
− | ==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}{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 <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 | -vsamc | ||
− | |||
+ | == 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}} |
Revision as of 10:54, 19 April 2021
Problem
Let and be the sequences of real numbers such that for all integers , where . What is
Solution 1
Square the given equality to yield so and
Solution 2 (DeMoivre's Formula)
Note that . Let , then, we know that , so . Therefore, . Aha is a geometric sequence that evaluates to . We can quickly see that . . Therefore, . The imaginary part is , so our answer is .
~AopsUser101, minor edit by vsamc stating that the answer choice is B, revamped by OreoChocolate
Solution 3
Clearly . So we have . By linearity, we have the latter is equivalent to . Expanding the summand yields -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.