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 "1964 AHSME Problems/Problem 31"

(Created page with "Let <cmath>f(n)=\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-3\sqrt{5}}{10}\left(\dfrac{1-\sqrt{5}}{2}\right)^n.</cmath> Then <math>f(n+1)-f(n-1)</math>,...")
 
(→‎Solution)
 
(5 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 +
==Problem==
 +
 
Let <cmath>f(n)=\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-3\sqrt{5}}{10}\left(\dfrac{1-\sqrt{5}}{2}\right)^n.</cmath>
 
Let <cmath>f(n)=\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-3\sqrt{5}}{10}\left(\dfrac{1-\sqrt{5}}{2}\right)^n.</cmath>
  
 
Then <math>f(n+1)-f(n-1)</math>, expressed in terms of <math>f(n)</math>, equals:
 
Then <math>f(n+1)-f(n-1)</math>, expressed in terms of <math>f(n)</math>, equals:
  
−
<math>\textbf{(A) }\frac{1}{2}f(n)\qquad\textbf{(B) }f(n)\qquad\textbf{(C) }2f(n)+1\qquad\textbf{(D) }f^2(n)\qquad \textbf{(E) }\frac{1}{2}(f^2(n)-1)</math>
+
<math>\textbf{(A) }\frac{1}{2}f(n)\qquad\textbf{(B) }f(n)\qquad\textbf{(C) }2f(n)+1\qquad\textbf{(D) }f^2(n)\qquad \textbf{(E) }</math>
 +
<math>\frac{1}{2}(f^2(n)-1)</math>
 +
 
 +
==Solution==
 +
We compute <math>f(n+1)</math> and <math>f(n-1)</math>, while pulling one copy of the exponential part outside:
 +
 
 +
<math>f(n+1) = \dfrac{5+3\sqrt{5}}{10}\cdot \frac{1 + \sqrt{5}}{2}\cdot\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-3\sqrt{5}}{10}\cdot \frac{1 - \sqrt{5}}{2}\cdot\left(\dfrac{1-\sqrt{5}}{2}\right)^n</math>
 +
 
 +
<math>f(n+1) = \dfrac{20+8\sqrt{5}}{20}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{20-8\sqrt{5}}{20}\left(\dfrac{1-\sqrt{5}}{2}\right)^n</math>
 +
 
 +
 
 +
 
 +
<math>f(n-1) = \dfrac{5+3\sqrt{5}}{10}\cdot \frac{2}{1 + \sqrt{5}}\cdot\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-3\sqrt{5}}{10}\cdot \frac{2}{1 - \sqrt{5}}\cdot\left(\dfrac{1-\sqrt{5}}{2}\right)^n</math>
 +
 
 +
<math>f(n-1) = \dfrac{(5+3\sqrt{5})(1 - \sqrt{5})}{5(1 + \sqrt{5})(1 - \sqrt{5})}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{(5-3\sqrt{5})(1 + \sqrt{5})}{5(1 - \sqrt{5})(1 + \sqrt{5})}\left(\dfrac{1-\sqrt{5}}{2}\right)^n</math>
 +
 
 +
<math>f(n-1) = \dfrac{-10-2\sqrt{5}}{5(-4)}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{-10 + 2\sqrt{5}}{5(-4)}\left(\dfrac{1-\sqrt{5}}{2}\right)^n</math>
 +
 
 +
<math>f(n-1) = \dfrac{10+2\sqrt{5}}{20}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{10 - 2\sqrt{5}}{20}\left(\dfrac{1-\sqrt{5}}{2}\right)^n</math>
 +
 
 +
 
 +
Computing <math>f(n+1) - f(n-1)</math> gives:
 +
 
 +
<math>f(n+1) - f(n-1) = \dfrac{10+6\sqrt{5}}{20}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{10-6\sqrt{5}}{20}\left(\dfrac{1-\sqrt{5}}{2}\right)^n</math>
 +
 
 +
<math>f(n+1) - f(n-1) = \dfrac{5+3\sqrt{5}}{10}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-3\sqrt{5}}{10}\left(\dfrac{1-\sqrt{5}}{2}\right)^n</math>
 +
 
 +
<math>f(n+1) - f(n-1) = f(n)</math>
 +
 
 +
Thus, the answer is <math>\boxed{\textbf{(B)}}</math>.
 +
 
 +
==Solution 2==
 +
Notice that <math>\left(\frac{1+\sqrt{5}}{2}\right)^n</math> and <math>\left(\frac{1-\sqrt{5}}{2}\right)^n</math> are the characteristics roots for the recurrence relation <math>F_n = F_{n-1} + F_{n-2}</math> (think about Binet's formula). And <math>f(n)</math> is the solution (i.e. <math>a_n</math>) to the recurrence relation with constants <math>a = \frac{5+3\sqrt{5}}{10}</math> and <math>b = \frac{5-3\sqrt{5}}{10}</math>. Thus, <math>f(n+1) - f(n-1) = f(n)</math>, and the answer is <math>\boxed{\textbf{(B)}}</math>. -nullptr07
 +
 
 +
==See Also==
 +
{{AHSME 40p box|year=1964|num-b=30|num-a=32}}
 +
 
 +
[[Category:Introductory Algebra Problems]]
 +
 
 +
{{MAA Notice}}

Latest revision as of 23:15, 29 June 2023

Problem

Let \[f(n)=\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-3\sqrt{5}}{10}\left(\dfrac{1-\sqrt{5}}{2}\right)^n.\]

Then $f(n+1)-f(n-1)$, expressed in terms of $f(n)$, equals:

$\textbf{(A) }\frac{1}{2}f(n)\qquad\textbf{(B) }f(n)\qquad\textbf{(C) }2f(n)+1\qquad\textbf{(D) }f^2(n)\qquad \textbf{(E) }$ $\frac{1}{2}(f^2(n)-1)$

Solution

We compute $f(n+1)$ and $f(n-1)$, while pulling one copy of the exponential part outside:

$f(n+1) = \dfrac{5+3\sqrt{5}}{10}\cdot \frac{1 + \sqrt{5}}{2}\cdot\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-3\sqrt{5}}{10}\cdot \frac{1 - \sqrt{5}}{2}\cdot\left(\dfrac{1-\sqrt{5}}{2}\right)^n$

$f(n+1) = \dfrac{20+8\sqrt{5}}{20}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{20-8\sqrt{5}}{20}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$


$f(n-1) = \dfrac{5+3\sqrt{5}}{10}\cdot \frac{2}{1 + \sqrt{5}}\cdot\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-3\sqrt{5}}{10}\cdot \frac{2}{1 - \sqrt{5}}\cdot\left(\dfrac{1-\sqrt{5}}{2}\right)^n$

$f(n-1) = \dfrac{(5+3\sqrt{5})(1 - \sqrt{5})}{5(1 + \sqrt{5})(1 - \sqrt{5})}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{(5-3\sqrt{5})(1 + \sqrt{5})}{5(1 - \sqrt{5})(1 + \sqrt{5})}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$

$f(n-1) = \dfrac{-10-2\sqrt{5}}{5(-4)}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{-10 + 2\sqrt{5}}{5(-4)}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$

$f(n-1) = \dfrac{10+2\sqrt{5}}{20}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{10 - 2\sqrt{5}}{20}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$


Computing $f(n+1) - f(n-1)$ gives:

$f(n+1) - f(n-1) = \dfrac{10+6\sqrt{5}}{20}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{10-6\sqrt{5}}{20}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$

$f(n+1) - f(n-1) = \dfrac{5+3\sqrt{5}}{10}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-3\sqrt{5}}{10}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$

$f(n+1) - f(n-1) = f(n)$

Thus, the answer is $\boxed{\textbf{(B)}}$.

Solution 2

Notice that $\left(\frac{1+\sqrt{5}}{2}\right)^n$ and $\left(\frac{1-\sqrt{5}}{2}\right)^n$ are the characteristics roots for the recurrence relation $F_n = F_{n-1} + F_{n-2}$ (think about Binet's formula). And $f(n)$ is the solution (i.e. $a_n$) to the recurrence relation with constants $a = \frac{5+3\sqrt{5}}{10}$ and $b = \frac{5-3\sqrt{5}}{10}$. Thus, $f(n+1) - f(n-1) = f(n)$, and the answer is $\boxed{\textbf{(B)}}$. -nullptr07

See Also

1964 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 30 		Followed by
Problem 32
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME 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=1964_AHSME_Problems/Problem_31&oldid=194976"