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.
The problems on this page are the property of the MAA's American Mathematics Competitions