Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3 (Characteristic Equation and Generic Formula))
(→‎Solution 3a (Generic Sum Formula))
Line 45: Line 45:
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
  
−
==Solution 3a (Generic Sum Formula) ==
+
==Solution 3a (Generic Lucas Sequence Sum Formula) ==
 
   
 
   
 
<cmath>
 
<cmath>
Line 73: Line 73:
 
<cmath>L_{12} = (L_{6} )^2 -2 = (L_{3}^2 + 2 )^2 -2  = (4^2 + 2 )^2 - 2  = 324 - 2 = 322  </cmath>
 
<cmath>L_{12} = (L_{6} )^2 -2 = (L_{3}^2 + 2 )^2 -2  = (4^2 + 2 )^2 - 2  = 324 - 2 = 322  </cmath>
  
−
Hence, <cmath>sum(L_{1} + L_{2} + ... + L_{10}) = L_{12} -3 = 322 - 3 =  \boxed{(B) 319} </cmath>.
+
Hence, <cmath>\sum_{n=1}^{10} = L_{12} -3 = 322 - 3 =  \boxed{(B) 319} </cmath>
 +
 
 +
Bonus Question: what is last digit of <math> L_{2^{2024}} </math> ?
  
 
==Solution 4==
 
==Solution 4==

Revision as of 17:46, 17 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 (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 leapfrog 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 Recurrence Relationship)

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

Solution 3a (Generic Lucas Sequence Sum Formula)

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

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: \[1 - A = B \quad \text{and} \quad 1 - B = A.\]

\[S_k = \frac{A(1 - A^k)}{B} + \frac{B(1 - B^k)}{A} = \frac{A^{k+2} - A^2 + B^{k+2} - B^2}{A \cdot B} = \frac{A^{k+2} + B^{k+2} - (A^2+ B^2) }{A \cdot B} = A^{k+2} + B^{k+2} -3 = L_{n+2} - 3\]

This formula $S_k = A^{k+2} + B^{k+2} - 3$ gives us a direct way to calculate the sum of $L_n$ for any $k$.

\[L_{2n} = (A^n + B^n)^2 - 2 \cdot (AB)^n = L_{n} - 2 \cdot (-1) ^n\] \[L_{2n+1} = (A^n + B^n)(A^{n+1} + B^{n+1}) - (AB)^n(A+B) = L_{n}L_{n+1} - (-1)^n\]

\[L_{12} = (L_{6} )^2 -2 = (L_{3}^2 + 2 )^2 -2 = (4^2 + 2 )^2 - 2 = 324 - 2 = 322\]

Hence, \[\sum_{n=1}^{10} = L_{12} -3 = 322 - 3 = \boxed{(B) 319}\]

Bonus Question: what is last digit of $L_{2^{2024}}$ ?

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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10B_Problems/Problem_23&oldid=234753"