Difference between revisions of "1970 AHSME Problems/Problem 16"
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\text{(E) } 26</math> | \text{(E) } 26</math> | ||
− | == | + | = 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 == | == See also == |
Latest revision as of 21:30, 13 July 2019
Contents
Problem
If is a function such that , and such that for then
Solution
Plugging in gives .
Plugging in gives .
Plugging in gives .
Thus, the answer is .
Sidenote
All the numbers in the sequence are integers. In fact, the function satisfies . (Prove it!).
See also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.