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Problem of the Month
Hint for Problem 0: September 2023

  1. f(r) is undefined only when r=2. For what value of r is f(r)=2?

  2. The sequence in part (i) is periodic. Can you show that the sequence is periodic for other values of r1?

    1. No hint given.

    2. After substituting the expression for f(r), multiply the numerator and denominator by r+1. Try to find a common factor in the numerator and denominator.

    3. Use (ii) and the fact that when r is positive, |12r+1|<12. Try to establish the given inequality for a few small values of n and observe how knowing the inequality for n can help you to deduce it for n+1.

  3. Three of these sequences are periodic, one of them is constant (after the first term), and one of them always approaches the fixed value 3132 as long as there are no undefined values in the sequence.