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

Revision as of 23:22, 24 May 2015

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:

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

Using the recursive definition, we find that $a_3=33$ .

All the terms in the sequence $a_n$ are integers. In fact, the sequence $a_n$ satisfies the recursion $a_n=3a_(n-1)+a_(n-2)$ .

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 39	Followed by Problem 41
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All AHSME Problems and Solutions

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 == Solution ==
 Using the recursive definition, we find that <math>a_3=33</math>.
+==Sidenote==
+All the terms in the sequence <math>a_n</math> are integers. In fact, the sequence <math>a_n</math> satisfies the recursion <math>a_n=3a_(n-1)+a_(n-2)</math>.
 == See Also ==