Difference between revisions of "2024 AMC 10B Problems/Problem 24"
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==Solution 1== | ==Solution 1== | ||
The first <math>20</math> terms <math>F_n = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765</math> | The first <math>20</math> terms <math>F_n = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765</math> | ||
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+ | so the answer is <math>1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = \boxed{(B) 319} </math>. | ||
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+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
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+ | ==Solution 2== | ||
+ | |||
+ | Define new sequence <cmath> G_n = \frac{F_{2n}}{F_{n}} = \frac{A^{2n} - B^{2n}}{A^{n} - B^{n}} =A^n+B^n </cmath> | ||
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+ | A= <math>\frac{1+\sqrt{5}}{2}</math> and B = <math>\frac{1-\sqrt{5}}{2}</math> | ||
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+ | Per characteristic equation, <math>G_n</math> itself is also Fibonacci type sequence with starting item <math>G_{1}=1 , G_{2}=3</math> | ||
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+ | then we can calculate the first 10 items using <math> G_{n} =G_{n-1} + G_{n-2} </math> | ||
so the answer is <math>1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = \boxed{(B) 319} </math>. | so the answer is <math>1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = \boxed{(B) 319} </math>. |
Revision as of 03:23, 14 November 2024
- The following problem is from both the 2024 AMC 10B #24 and 2024 AMC 12B #18, so both problems redirect to this page.
Contents
[hide]Problem 18
The Fibonacci numbers are defined by and for What is
Solution 1
The first terms
so the answer is .
Solution 2
Define new sequence
A= and B =
Per characteristic equation, itself is also Fibonacci type sequence with starting item
then we can calculate the first 10 items using
so the answer is .
See also
2024 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 10 Problems and Solutions |
2024 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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.