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

Revision as of 07:29, 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.

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

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

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

Solution 3

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

Solution 4

Remember that for any $n>=0$ , $\frac{F_{2n}}{F_{n}} = L_{n}$

Therefore, the problem can be expressed as the sum of the first 10 Lucas numbers, starting at 1,

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

~Apollo08 (first solution)

@@ Line 27: / Line 27: @@
 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>.
 ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
+==Solution 4==
+Remember that for any <math> n>=0 </math>, <cmath> \frac{F_{2n}}{F_{n}} = L_{n} </cmath>
+Therefore, the problem can be expressed as the sum of the first 10 Lucas numbers, starting at 1,
+making the answer <math> 1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = \boxed{(B) 319} </math>.
+~Apollo08 (first solution)
 ==See also==
 {{AMC10 box|year=2024|ab=B|num-b=22|num-a=24}}
 {{MAA Notice}}

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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