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Difference between revisions of "2024 AMC 10B Problems/Problem 24"

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{{duplicate|[[2024 AMC 10B Problems/Problem 24|2024 AMC 10B #24]] and [[2024 AMC 12B Problems/Problem 18|2024 AMC 12B #18]]}}
 
{{duplicate|[[2024 AMC 10B Problems/Problem 24|2024 AMC 10B #24]] and [[2024 AMC 12B Problems/Problem 18|2024 AMC 12B #18]]}}
  
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==Problem==
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==Problem 18==
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The Fibonacci numbers are defined by <math>F_1=1,</math> <math>F_2=1,</math> and <math>F_n=F_{n-1}+F_{n-2}</math> for <math>n\geq 3.</math> What is<cmath>\dfrac{F_2}{F_1}+\dfrac{F_4}{F_2}+\dfrac{F_6}{F_3}+\cdots+\dfrac{F_{20}}{F_{10}}?</cmath>
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<math>\textbf{(A) }318 \qquad\textbf{(B) }319\qquad\textbf{(C) }320\qquad\textbf{(D) }321\qquad\textbf{(E) }322</math>
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==Solution #1==
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1,1,2,3,5,8,13,21, 34, 55, 89, 144, 233, 377, 610,  987, 1597, 2584, 4181, 6765
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so ask = 1+  3 + 4 + 7 + 11 + 18 + +29 +47 + 76 + 123 =  <math>\boxed{B 319} </math>.
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~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
  
 
==See also==
 
==See also==

Revision as of 02:43, 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 $F_1=1,$ $F_2=1,$ and $F_n=F_{n-1}+F_{n-2}$ for $n\geq 3.$ What is\[\dfrac{F_2}{F_1}+\dfrac{F_4}{F_2}+\dfrac{F_6}{F_3}+\cdots+\dfrac{F_{20}}{F_{10}}?\] $\textbf{(A) }318 \qquad\textbf{(B) }319\qquad\textbf{(C) }320\qquad\textbf{(D) }321\qquad\textbf{(E) }322$

Solution #1

1,1,2,3,5,8,13,21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765

so ask = 1+ 3 + 4 + 7 + 11 + 18 + +29 +47 + 76 + 123 = $\boxed{B 319}$.

~luckuso

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. AMC logo.png

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