Problem 23 · 2017 Math Kangaroo
Stretch
Algebra & Patterns
arithmetic-sequence
We look at the sequence \(\langle a_n \rangle\) with \(a_1 = 2017\) and \(a_{n+1} = \dfrac{a_n - 1}{a_n}\). Then \(a_{999} =\)
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Answer: E — \(-\dfrac{1}{2016}\)
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Hint 1 of 2
Compute the first few terms; recurrences like this often repeat with a short period.
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Hint 2 of 2
Find the period, then use 999 modulo that period to locate a_999.
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Approach: detect the period of the recurrence
- a1 = 2017, a2 = 2016/2017, a3 = (a2 - 1)/a2 = -1/2016, a4 = (a3 - 1)/a3 = 2017 = a1.
- So the sequence repeats with period 3.
- 999 is a multiple of 3, so a_999 = a_3 = -1/2016.
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