SOLUTION: 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

Question 1064172: 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 by Edwin McCravy(20063) (Show Source): You can put this solution on YOUR website!


We calculate enough terms to discover the pattern,
Here are the first dozen terms:

 a(1) = 2001
 a(2) = 2002
 a(3) = 2003
 a(4) = 2001+2002-2003 = 2000
 a(5) = 2002+2003-2000 = 2005
 a(6) = 2003+2000-2005 = 1998
 a(7) = 2000+2005-1998 = 2007
 a(8) = 2005+1998-2007 = 1996
 a(9) = 1998+2007-1996 = 2009
a(10) = 2007+1996-2009 = 1994
a(11) = 1996+2009-1994 = 2011
a(12) = 2009+1994-2011 = 1992
...

We want to find a(2004)
That's an even term. We look at the even terms:

 n   a(n)
 --------
 2   2002
 4   2000
 6   1998
 8   1996
10   1994
12   1992
 
We see that the pattern for the even terms is 
"if you add n and a(n), the sum is always 2004."

That is n + a(n) = 2004, or,

            a(n) = 2004-n when n is even

So when n = 2004, a(2004) = 2004-2004 = 0

Answer: 0

Edwin

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