1988 AHSME Problems/Problem 30
Problem
Let . Give , consider the sequence defined by for all . For how many real numbers will the sequence take on only a finite number of different values?
Solution
Note that gives the constant sequence , since . Because gives the sequence with two different values. Similarly, , so gives the sequence with three values. In general, if gives the sequence with different values, and , then gives a sequence with different values. (It is easy to see that we could not have for some .) Thus, it follows by induction that there is a sequence with distinct values for every positive integer , as long as we can verify that there is always a real number such that . This makes the answer . The verification alluded to above, which completes the proof, follows from the quadratic formula: the solutions to are . Hence if , then is real, since the part under the square root is non-negative, and in fact , since will be between and , so the square root will be between and , and something between and gives something between and . Finally, since , it follows by induction that all terms satisfy ; in particular, they are all real.
See also
1988 AHSME (Problems • Answer Key • Resources) | ||
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