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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==
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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.

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

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 the answer is $1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = \boxed{(B) 319}$.

~luckuso

Solution 2

Define new sequence \[G_n = \frac{F_{2n}}{F_{n}} = \frac{A^{2n} - B^{2n}}{A^{n} - B^{n}} =A^n+B^n\]

A= $\frac{1+\sqrt{5}}{2}$ and B = $\frac{1-\sqrt{5}}{2}$

Per characteristic equation, $G_n$ itself is also Fibonacci type sequence with starting item $G_{1}=1 , G_{2}=3$

then we can calculate the first 10 items using $G_{n} =G_{n-1} + G_{n-2}$

so the answer is $1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = \boxed{(B) 319}$.

~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

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