Difference between revisions of "1964 AHSME Problems/Problem 31"
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<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>\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> | <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
Contents
Problem
Let
Then , expressed in terms of
, equals:
Solution
We compute and
, while pulling one copy of the exponential part outside:
Computing gives:
Thus, the answer is .
Solution 2
Notice that and
are the characteristics roots for the recurrence relation
(think about Binet's formula). And
is the solution (i.e.
) to the recurrence relation with constants
and
. Thus,
, and the answer is
. -nullptr07
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 30 |
Followed by Problem 32 | |
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All AHSME Problems and Solutions |
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