1958 AHSME Problems/Problem 40

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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

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

Sidenote

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

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 39
Followed by
Problem 41
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