2004 AMC 12B Problem 12

Attempt Problem 12 of the 2004 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 12B solutions, or check the answer key.

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12.

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?
−2004-2004
−2-2
00
40034003
60076007
Answer: C
Concepts:recursionarithmetic sequencepattern recognition
Small hint:
Write out several terms to spot a pattern
Big hint:
The even-indexed terms 2002,2000,1998,…2002, 2000, 1998, \ldots form an arithmetic progression
Solution:
The rule gives 2001,2001, 2002,2002, 2003,2003, 2000,2000, 2005,2005, 1998,…1998, \ldots The even-indexed terms are 2002,2000,1998,…,2002, 2000, 1998, \ldots, decreasing by 2.2.
More precisely, the recurrence verifies inductively that a2k=2004−2ka_{2k} = 2004 - 2k and a2k+1=2001+2k.a_{2k+1} = 2001 + 2k. Therefore a2004=a2⋅1002=2004−2(1002)=0. \begin{aligned} a_{2004} &= a_{2\cdot1002} \\ &= 2004 - 2(1002) = 0. \end{aligned}
Thus, the correct answer is C.
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