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Difference between revisions of "2024 AMC 10B Problems/Problem 23"
(→‎Solution 3a (Generic Lucas Sequence Sum Formula))
 
(23 intermediate revisions by 6 users not shown)
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==Solution 1 (Bash)==
 
==Solution 1 (Bash)==
−
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>
+
The first <math>20</math> terms are <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>.  
+
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]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
  
 
==Solution 2 (Pattern)==
 
==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 <math>{\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,</math> and <math>{\frac{F_8}{F_4}} = {\frac{21}{3}} = 7.</math> The pattern is that then ten fractions are in their own Fibonacci sequence with the starting two terms being <math>1</math> and <math>3</math>, which can be written as <math>G_1 = 1, G_2 = 3, G_n = G_{n-1} + G_{n-2}</math> for <math>n \geq 3.</math> The problem is asking for the sum of the ten terms <math>G_1 + G_2 + G_3 + ... + G_{10}</math>, and after adding up the first ten terms <math>1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123</math>, you arrive at the solution <math>\boxed{\textbf{(B) }319}.</math>
+
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 <math>{\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,</math> and <math>{\frac{F_8}{F_4}} = {\frac{21}{3}} = 7.</math> The pattern is that then ten fractions are in their own leapfrog sequence with the starting two terms being <math>1</math> and <math>3</math>, which can be written as <math>G_1 = 1, G_2 = 3, G_n = G_{n-1} + G_{n-2}</math> for <math>n \geq 3.</math> The problem is asking for the sum of the ten terms <math>G_1 + G_2 + G_3 + ... + G_{10}</math>, and after adding up the first ten terms <math>1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123</math>, you arrive at the solution <math>\boxed{\textbf{(B) }319}.</math>
  
 
~Cattycute
 
~Cattycute
  
−
==Solution 3 (Characteristic Equation) ==
+
Note: The first ten terms (<math>1+3+4+7+...+76+123</math>) are actually part of a sequence called the Lucas numbers.
  
−
Define new sequence <cmath>  G_n = \frac{F_{2n}}{F_{n}} = \frac{A^{2n} - B^{2n}}{A^{n} - B^{n}} =A^n+B^n </cmath>
+
~NXC
  
−
A= <math>\frac{1+\sqrt{5}}{2}</math> and B = <math>\frac{1-\sqrt{5}}{2}</math>
+
==Solution 3 (Characteristic Equation and Recurrence Relationship) ==
  
−
Per characteristic equation, <math>G_n</math> itself is also Fibonacci type sequence with starting item <math>G_{1}=1 , G_{2}=3</math>  
+
Using characteristic equation <math>x^2=x+1</math> and starting terms <math>F_1 =1, F_2=1</math>, we get <math>F_n=\frac{1}{\sqrt{5}} (A^n- B^n) </math>, A= <math>\frac{1+\sqrt{5}}{2}</math> , B = <math>\frac{1-\sqrt{5}}{2}</math>  
  
−
then we can calculate the first 10 items using <math>  G_{n}  =G_{n-1}  + G_{n-2} </math>
+
Define a new sequence (this is actually a Lucas sequence) <cmath>  L_n = \frac{F_{2n}}{F_{n}} = \frac{A^{2n} - B^{2n}}{A^{n} - B^{n}} =A^n+B^n </cmath>
 +
 
 +
Given that <math>L_n</math> has the same 2 roots (A,B) as the characteristic equation <math>x^2=x+1</math> it implies that <math>  L_{n}  =L_{n-1}  + L_{n-2} </math>. It can also be verified below.
 +
Noticing <math> A^2 = A + 1 </math> and <math> B^2 = B + 1 </math>,
 +
Therefore,
 +
<cmath>
 +
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} .
 +
</cmath>
 +
 
 +
Calculating the first 10 terms using the recurrence relation <math>  L_{n}  =L_{n-1}  + L_{n-2} </math> and the initial values <math> L_1 = 1 </math> and <math> L_2 = 3 </math>, we get:
 +
<cmath>
 +
L_1 = 1, \quad L_2 = 3, \quad L_3 = 4, \quad L_4 = 7, \quad L_5 = 11 </cmath>
 +
 
 +
<cmath>\quad L_6 = 18, \quad L_7 = 29, \quad L_8 = 47, \quad L_9 = 76, \quad L_{10} = 123
 +
</cmath>
 +
 
 +
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 3a (Generic Lucas Sequence Sum Formula) ==
 +
 +
<cmath>
 +
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}
 +
</cmath>
 +
 
 +
Recall the sum formula for a geometric series:
 +
<cmath>
 +
\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
 +
</cmath>
 +
 
 +
To simplify further, notice that:
 +
<math> A = \frac{1 + \sqrt{5}}{2} </math> and <math> B = \frac{1 - \sqrt{5}}{2} </math> are roots of <math> x^2 = x + 1 </math>, so:
 +
<cmath>
 +
1 - A = B \quad \text{and} \quad 1 - B = A.
 +
</cmath>
 +
 
 +
<cmath>
 +
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
 +
</cmath>
 +
 
 +
This formula <math>S_k =  A^{k+2}  + B^{k+2} - 3 </math> gives us a direct way to calculate the sum of <math> L_n </math> for any <math> k </math>.
 +
 
 +
<cmath>L_{2n} = (A^n + B^n)^2 - 2 \cdot (AB)^n = L_{n} - 2 \cdot (-1) ^n</cmath>
 +
<cmath>L_{2n+1} = (A^n + B^n)(A^{n+1} + B^{n+1}) -  (AB)^n(A+B) = L_{n}L_{n+1} - (-1)^n</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_{n=1}^{10} = L_{12} -3 = 322 - 3 =  \boxed{(B) 319} </cmath>
  
−
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]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
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Remember that for any <math> n\ge0 </math>, <cmath> \frac{F_{2n}}{F_{n}} = L_{n} </cmath>
 
Remember that for any <math> n\ge0 </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,
+
so the answer is <math> 1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = \boxed{(B) 319} </math>.
−
 
 
−
making the answer <math> 1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = \boxed{(B) 319} </math>.
 
  
 
~Apollo08 (first solution)
 
~Apollo08 (first solution)
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https://www.youtube.com/watch?v=7KEk5VbxwAU
 
https://www.youtube.com/watch?v=7KEk5VbxwAU
  
 +
==Video Solution 3 by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=24EZaeAThuE
  
 
==See also==
 
==See also==

Latest revision as of 20:37, 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 are $F_n = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765$

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

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

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


~luckuso

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

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