Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2024 AMC 10B Problems/Problem 24: Difference between revisions

← Older editNewer edit →
Bloggish (talk | contribs)
editor
76 edits
No edit summary
Bloggish (talk | contribs)
editor
76 edits
Undo revision 233818 by Bloggish (talk)
Tag: Undo
Line 1: Line 1:
{{duplicate|[[2024 AMC 10B Problems/Problem 24|2024 AMC 10B #24]] and [[2024 AMC 12B Problems/Problem 18|2024 AMC 12B #18]]}}
==Problem 18==
==Problem 18==
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>
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>

Revision as of 05:50, 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10B_Problems/Problem_24&oldid=233819"