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標題:
Maths Problem(different values in sequence)
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最佳解答:
From the given of the question, for any positive integer n, we have: an+1 = an an+2 - 1 Rearranging, we have: an+2 = (an+1 + 1)/an with a1 = a, a2 = 2000 So, 圖片參考：http://www.geocities.com/abcsghk/7006121001797.gif So, the sequence is periodic and a1 = a6 = a11 = ... For 2001 to appear in this sequence, either a = 2001 ----------(1) 2000 = 2001 ----------(2) 2001/a = 2001 -----------(3) (2001+a)/(2000a) = 2001 ---------(4) (a+1)/2000 = 2001 -----------(5) For (1), a = 2001 For (2), no solution For (3), 2001/a = 2001 → a = 1 For (4), (2001+a)/(2000a) = 2001 → 2001+a = 4002000a → a = 2001/4001999 For (5), (a+1)/2000 = 2001 → a+1 = 4002000 → a = 4001999 In conclusion, if a = 2001/4001999, a = 1, a = 2001, a = 4001999, then 2001 will appear in the sequence. That is, there are 4 such values of a. Hope it helps! ^^
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Maths Problem(different values in sequence)
發問:
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a, 2000, b, ..., is a sequence of real number > 0. Each term after the first is 1 less than the product of its two adjacent terms. For how many different values of a does 2001 appear in the sequence?最佳解答:
From the given of the question, for any positive integer n, we have: an+1 = an an+2 - 1 Rearranging, we have: an+2 = (an+1 + 1)/an with a1 = a, a2 = 2000 So, 圖片參考：http://www.geocities.com/abcsghk/7006121001797.gif So, the sequence is periodic and a1 = a6 = a11 = ... For 2001 to appear in this sequence, either a = 2001 ----------(1) 2000 = 2001 ----------(2) 2001/a = 2001 -----------(3) (2001+a)/(2000a) = 2001 ---------(4) (a+1)/2000 = 2001 -----------(5) For (1), a = 2001 For (2), no solution For (3), 2001/a = 2001 → a = 1 For (4), (2001+a)/(2000a) = 2001 → 2001+a = 4002000a → a = 2001/4001999 For (5), (a+1)/2000 = 2001 → a+1 = 4002000 → a = 4001999 In conclusion, if a = 2001/4001999, a = 1, a = 2001, a = 4001999, then 2001 will appear in the sequence. That is, there are 4 such values of a. Hope it helps! ^^
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