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

Latest revision as of 22:44, 8 March 2024

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 $a_n=3a_{n-1}+a_{n-2}$ (Prove it!).

1958 AHSC (Problems • Answer Key • Resources)
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

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 == 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
@@ Line 6: / Line 6: @@
 \textbf{(C)}\ 21\qquad
 \textbf{(D)}\ 10\qquad
-\textbf{(E)}\ \minus{}17</math>
+\textbf{(E)}\ -17</math>
 == Solution ==
-<math>\fbox{}</math>
+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> (Prove it!).
 == See Also ==