2024 AMC 10B Problems/Problem 23

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 (Bash)

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 (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 ${\frac{F_2}{F_1}} = {\frac{1}{1}} = 1, {\frac{F_4}{F_2}} = {\frac{3}{1}} = 3, {\frac{F_6}{F_3}} = {\frac{8}{2}} = 4,$ 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 after adding up the first ten terms $1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123$ , you arrive at the solution $\boxed{\textbf{(B) }319}.$

~Cattycute

Note: The first ten terms ( $1+3+4+7+...+76+123$ ) are actually part of a sequence called the Lucas numbers.

~NXC

Solution 3 (Characteristic Equation and Generic Formula)

Using characteristic equation $x^2=x+1$ and starting terms $F_1 =1, F_2=1$ , we get $F_n=\frac{1}{\sqrt{5}} (A^n- B^n)$ , A= $\frac{1+\sqrt{5}}{2}$ , B = $\frac{1-\sqrt{5}}{2}$

Define new sequence (this is actually Lucas sequence) $L_n = \frac{F_{2n}}{F_{n}} = \frac{A^{2n} - B^{2n}}{A^{n} - B^{n}} =A^n+B^n$

Given $L_n$ has same 2 roots A , B from characteristic equation $x^2=x+1$ which implies $L_{n} =L_{n-1} + L_{n-2}$ , it can also be verified below. Noticing $A^2 = A + 1$ and $B^2 = B + 1$ , Therefore, $L_n = A^n + B^n = A^2 \cdot A^{n-2} + B^2 \cdot B^{n-2} = (A +1) \cdot A^{n-2} + (B+1) \cdot B^{n-2} = (A^{n-1} + B^{n-1}) + (A^{n-2} + B^{n-2}) = L_{n-1} + L_{n-2} .$

Calculate the first 10 terms using the recurrence relation $L_{n} =L_{n-1} + L_{n-2}$ and the initial values $L_1 = 1$ and $L_2 = 3$ , we get: $L_1 = 1, \quad L_2 = 3, \quad L_3 = 4, \quad L_4 = 7, \quad L_5 = 11$

$\quad L_6 = 18, \quad L_7 = 29, \quad L_8 = 47, \quad L_9 = 76, \quad L_{10} = 123$

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

~luckuso

Appendix : Closed-form expression for the sum of the Lucas sequence $L_n$ ,

$S_k = \sum_{n=1}^{k} L_n = \sum_{n=1}^{k} (A^n + B^n). = \sum_{n=1}^{k} A^n + \sum_{n=1}^{k} B^n = \frac{A(1 - A^k)}{1 - A} + \frac{B(1 - B^k)}{1 - B}.$

Recall the sum formula for a geometric series: $\sum_{n=1}^{k} r^n = r + r^2 + \cdots + r^k = \frac{r(1 - r^k)}{1 - r}, \quad \text{for } r \neq 1.$

To simplify further, notice that: - $A = \frac{1 + \sqrt{5}}{2}$ and $B = \frac{1 - \sqrt{5}}{2}$ are roots of $x^2 = x + 1$ , so: $1 - A = -B \quad \text{and} \quad 1 - B = -A.$

$S_k = \frac{A(1 - A^k)}{-B} + \frac{B(1 - B^k)}{-A} = \frac{A^{k+1} - A - B^{k+1} + B}{\sqrt{5}}.$ This formula gives us a direct way to calculate the sum of $L_n$ for any $k$ .

Solution 4

Remember that for any $n\ge0$ , $\frac{F_{2n}}{F_{n}} = L_{n}$

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

~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

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

Page

Toolbox

Search

2024 AMC 10B Problems/Problem 23

Contents

Problem

Solution 1 (Bash)

Solution 2 (Pattern)

Solution 3 (Characteristic Equation and Generic Formula)

Solution 4

Video Solution 1 by Pi Academy (In Less Than 3 Mins ⚡🚀)

Video Solution 2 by Innovative Minds

Video Solution 3 by SpreadTheMathLove

See also