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

## Problem

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 $F(6)=$

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

# Solution

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 $\fbox{C}$.

## Sidenote

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!).

## See also

 1970 AHSC (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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