Difference between revisions of "2024 AMC 10B Problems/Problem 24"
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The first 20 terms F_n = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765 | The first 20 terms F_n = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765 | ||
| − | so ans = 1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = <math>\boxed{B | + | so ans = 1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = <math>\boxed{(B) 319} </math>. |
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
Revision as of 02:53, 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.
Problem 18
The Fibonacci numbers are defined by 
and 
for 
What is![\[\dfrac{F_2}{F_1}+\dfrac{F_4}{F_2}+\dfrac{F_6}{F_3}+\cdots+\dfrac{F_{20}}{F_{10}}?\]](http://latex.artofproblemsolving.com/e/c/7/ec74725a3dc8fad2ea4f6227370af4217e3b0440.png)
Solution #1
The first 20 terms F_n = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765
so ans = 1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = 
.
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. 
