Difference between revisions of "2024 AMC 10B Problems/Problem 23"
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− | Note: The first ten terms (<math>1+3+4+7...76+123</math>) are actually part of a sequence called the Lucas numbers. | + | Note: The first ten terms (<math>1+3+4+7+...+76+123</math>) are actually part of a sequence called the Lucas numbers. |
~NXC | ~NXC |
Revision as of 20:35, 14 November 2024
- The following problem is from both the 2024 AMC 10B #23 and 2024 AMC 12B #18, so both problems redirect to this page.
Contents
[hide]Problem
The Fibonacci numbers are defined by and for What is
Solution 1 (Bash)
The first terms
so the answer is .
Solution 2 (Pattern)
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 after adding up the first ten terms , you arrive at the solution
~Cattycute
Note: The first ten terms () are actually part of a sequence called the Lucas numbers.
~NXC
Solution 3 (Characteristic Equation)
Define new sequence
A= and B =
Per characteristic equation, itself is also Fibonacci type sequence with starting item
then we can calculate the first 10 items using
so the answer is .
Solution 4
Remember that for any ,
so the answer is .
~Apollo08 (first solution)
Video Solution 1 by Pi Academy (In Less Than 3 Mins ⚡🚀)
https://youtu.be/YjQ9Q93RCu4?feature=shared
~ Pi Academy
Video Solution 2 by Innovative Minds
https://www.youtube.com/watch?v=7KEk5VbxwAU
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=24EZaeAThuE
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 |
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.