1958 AHSME Problems/Problem 40

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
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 41 42 43 44 45 46 47 48 49 50
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