Problem:
In the sequence 2001, 2002, 2003, … , 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. What is the 2004th term in this sequence?
Answer Choices:
A. −2004
B. −2
C. 0
D. 4003
E. 6007
Solution:
Let ak​ be the kth term of the sequence. For k≥3,
ak+1​=ak−2​+ak−1​−ak​, so ak+1​−ak−1​=−(ak​−ak−2​).
Because the sequence begins
2001,2002,2003,2000,2005,1998,…
it follows that the odd-numbered terms and the even-numbered terms each form arithmetic progressions with common differences of 2 and −2, respectively. The 2004th term of the original sequence is the 1002nd term of the sequence 2002,2000,1998,…, and that term is 2002+1001(−2)=0.
Answer: C​.
The problems on this page are the property of the MAA's American Mathematics Competitions