Difference between revisions of "2020 AMC 12A Problems/Problem 22"
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~AopsUser101, minor edit by vsamc stating that the answer choice is B | ~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}{ | + | 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 <math>\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)}</math> |
-vsamc | -vsamc | ||
==See Also== | ==See Also== |
Latest revision as of 16:20, 13 March 2020
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 .(Which is answer choice
~AopsUser101, minor edit by vsamc stating that the answer choice is B
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 |
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