Difference between revisions of "2024 AMC 10B Problems/Problem 23"
(→Solution 1) |
|||
| Line 6: | Line 6: | ||
==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> | |
| + | |||
| + | so the answer is <math>1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = \boxed{(B) 319} </math>. | ||
| + | |||
| + | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 05:40, 14 November 2024
Possible duplicate of https://artofproblemsolving.com/wiki/index.php/2024_AMC_10B_Problems/Problem_24
Contents
Problem
The Fibonacci numbers are defined by 
and 
for 
What is ![\[{\frac{F_2}{F_1}} + {\frac{F_4}{F_2}} + {\frac{F_6}{F_3}} + ... + {\frac{F_{20}}{F_{10}}}?\]](http://latex.artofproblemsolving.com/9/b/1/9b131758b581e3494845bf6cd1444b190f2deb1d.png)
Solution 1
The first 
terms 
so the answer is 
.
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 
and 
The pattern is that then ten fractions are in their own Fibonacci sequence with the starting two terms being 
and 
, which can be written as 
for The problem is asking for the sum of the ten terms 
, and you arrive at the solution 
~Cattycute
See also
| 2024 AMC 10B (Problems • Answer Key • Resources) | ||
| 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. 
