Difference between revisions of "1958 AHSME Problems/Problem 40"

(Created page with "== Problem == Given <math> a_0 \equal{} 1</math>, <math> a_1 \equal{} 3</math>, and the general relation <math> a_n^2 \minus{} a_{n \minus{} 1}a_{n \plus{} 1} \equal{} (\minus{}1...")
 
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== Problem ==
 
== Problem ==
Given <math> a_0 \equal{} 1</math>, <math> a_1 \equal{} 3</math>, and the general relation <math> a_n^2 \minus{} a_{n \minus{} 1}a_{n \plus{} 1} \equal{} (\minus{}1)^n</math> for <math> n \ge 1</math>. Then <math> a_3</math> equals:
+
Given <math> a_0 = 1</math>, <math> a_1 = 3</math>, and the general relation <math> a_n^2 - a_{n - 1}a_{n + 1} = (-1)^n</math> for <math> n \ge 1</math>. Then <math> a_3</math> equals:
  
 
<math> \textbf{(A)}\ \frac{13}{27}\qquad  
 
<math> \textbf{(A)}\ \frac{13}{27}\qquad  
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\textbf{(C)}\ 21\qquad  
 
\textbf{(C)}\ 21\qquad  
 
\textbf{(D)}\ 10\qquad  
 
\textbf{(D)}\ 10\qquad  
\textbf{(E)}\ \minus{}17</math>
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\textbf{(E)}\ -17</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 23:25, 13 March 2015

Problem

Given $a_0 = 1$, $a_1 = 3$, and the general relation $a_n^2 - a_{n - 1}a_{n + 1} = (-1)^n$ for $n \ge 1$. Then $a_3$ equals:

$\textbf{(A)}\ \frac{13}{27}\qquad  \textbf{(B)}\ 33\qquad  \textbf{(C)}\ 21\qquad  \textbf{(D)}\ 10\qquad  \textbf{(E)}\ -17$

Solution

$\fbox{}$

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 39
Followed by
Problem 41
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All AHSME Problems and Solutions

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