Difference between revisions of "1970 AHSME Problems/Problem 16"

Latest revision as of 12:39, 16 July 2024

If $F(n)$ is a function such that $F(1)=F(2)=F(3)=1$ , and such that $F(n+1)= \frac{F(n)\cdot F(n-1)+1}{F(n-2)}$ for $n\ge 3,$ then $F(6)=$

$\text{(A) } 2\quad \text{(B) } 3\quad \text{(C) } 7\quad \text{(D) } 11\quad \text{(E) } 26$

Plugging in $n=3$ gives $F(4) = \frac{F(3) \cdot F(2) + 1}{F(1)} = \frac{1 \cdot 1 + 1}{1} = 2$ .

Plugging in $n=4$ gives $F(5) = \frac{F(4) \cdot F(3) + 1}{F(2)} = \frac{2 \cdot 1 + 1}{1} = 3$ .

Plugging in $n=5$ gives $F(6) = \frac{F(5) \cdot F(4) + 1}{F(3)} = \frac{3 \cdot 2 + 1}{1} = 7$ .

Thus, the answer is $\fbox{C}$ .

All the numbers in the sequence $F(n)$ are integers. In fact, the function $F$ satisfies $F(n)=4F(n-2)-F(n-4)$ . (Prove it!).

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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
All AHSME Problems and Solutions

@@ Line 11: / Line 11: @@
 == Solution ==
-<math>\fbox{C}</math>
+Plugging in <math>n=3</math> gives <math>F(4) = \frac{F(3) \cdot F(2) + 1}{F(1)} = \frac{1 \cdot 1 + 1}{1} = 2</math>.
+Plugging in <math>n=4</math> gives <math>F(5) = \frac{F(4) \cdot F(3) + 1}{F(2)} = \frac{2 \cdot 1 + 1}{1} = 3</math>.
+Plugging in <math>n=5</math> gives <math>F(6) = \frac{F(5) \cdot F(4) + 1}{F(3)} = \frac{3 \cdot 2 + 1}{1} = 7</math>.
+Thus, the answer is <math>\fbox{C}</math>.
+==Sidenote==
+All the numbers in the sequence <math>F(n)</math> are integers. In fact, the function <math>F</math> satisfies <math>F(n)=4F(n-2)-F(n-4)</math>. (Prove it!).
 == See also ==
-{{AHSME box|year=1970|num-b=15|num-a=17}}
+{{AHSME 35p box|year=1970|num-b=15|num-a=17}}
 [[Category: Introductory Algebra Problems]]
 {{MAA Notice}}