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2004 AMC 10B Problem 19

Problem 19 of 25HarderAlgebra

In the sequence 2001,2001, 2002,2002, 2003,2003, …,\ldots, each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is 2001+2002−2003=2000.2001 + 2002 - 2003 = 2000. What is the 20042004th term in this sequence?

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Solution

The recurrence ak+1=ak−2+ak−1−aka_{k+1} = a_{k-2} + a_{k-1} - a_k gives ak+1−ak−1=−(ak−ak−2).a_{k+1} - a_{k-1} = -(a_k - a_{k-2}). The sequence begins 2001,2001, 2002,2002, 2003,2003, 2000,2000, 2005,2005, 1998,1998, …\ldots So the even-position terms form the arithmetic sequence 2002,2000,1998,…2002, 2000, 1998, \ldots with common difference −2.-2. The 20042004th term is its 10021002nd term, 2002+1001(−2)=0.2002 + 1001(-2) = 0. Thus, the correct answer is C.

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Concepts: recursion · arithmetic sequence · pattern recognition

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