Difference between revisions of "2024 AMC 10B Problems/Problem 23"

(Problem)
(Problem)
Line 1: Line 1:
 +
Possible duplicate [https://artofproblemsolving.com/wiki/index.php/2024_AMC_10B_Problems/Problem_24]
 +
 
==Problem==
 
==Problem==
 
The Fibonacci numbers are defined by <math>F_1 = 1, F_2 = 1,</math> and <math>F_n = F_{n-1} + F_{n-2}</math> for <math>n \geq 3.</math> What is <cmath>{\frac{F_2}{F_1}} + {\frac{F_4}{F_2}} + {\frac{F_6}{F_3}} + ... + {\frac{F_{20}}{F_{10}}}?</cmath>
 
The Fibonacci numbers are defined by <math>F_1 = 1, F_2 = 1,</math> and <math>F_n = F_{n-1} + F_{n-2}</math> for <math>n \geq 3.</math> What is <cmath>{\frac{F_2}{F_1}} + {\frac{F_4}{F_2}} + {\frac{F_6}{F_3}} + ... + {\frac{F_{20}}{F_{10}}}?</cmath>

Revision as of 04:39, 14 November 2024

Possible duplicate [1]

Problem

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 \[{\frac{F_2}{F_1}} + {\frac{F_4}{F_2}} + {\frac{F_6}{F_3}} + ... + {\frac{F_{20}}{F_{10}}}?\] $\textbf{(A) } 318 \qquad\textbf{(B) } 319 \qquad\textbf{(C) } 320 \qquad\textbf{(D) } 321 \qquad\textbf{(E) } 322$

Solution 1

Brute forcing gets you B) 319

Solution 2

Plug in a few numbers to see if there is a pattern. List out a few Fibonacci numbers, and then try them on the equation. You'll find that ${\frac{F_2}{F_1}} = {\frac{1}{1}} = 1, {\frac{F_4}{F_2}} = {\frac{3}{1}} = 3, {\frac{F_6}{F_3}} = {\frac{8}{2}} = 4,$ and ${\frac{F_8}{F_4}} = {\frac{21}{3}} = 7.$ The pattern is that then ten fractions are in their own Fibonacci sequence with the starting two terms being $1$ and $3$, which can be written as $G_1 = 1, G_2 = 3, G_n = G_{n-1} + G_{n-2}$ for $n \geq 3.$ The problem is asking for the sum of the ten terms $G_1 + G_2 + G_3 + ... + G_{10}$, and you arrive at the solution $\boxed{\textbf{(B) }319}.$

~Cattycute

See also

2024 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png