2004 AMC 10B Problem 19

P16-20~4 minPrint

Sequences & Series Use an invariant, parity, or coloring argument Compute small cases, spot the pattern, generalize

In the sequence $2001$ , $2002$ , $2003$ , $\ldots$ , 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 $2001 + 2002 - 2003 = 2000$ . What is the $2004^\textrm{th}$ term in this sequence?

Show solution without answering

Key insight

List the first ten terms. The even-numbered terms drop by 2 each time: 2002, 2000, 1998, ... After 1001 drops, term 2004 is 0.

Solution

The rule is "add the two terms before the last one, then subtract the last one." It is hard to see what term $2004$ is directly, so compute some terms and look for a pattern.

Compute the first ten terms:

Position	1	2	3	4	5	6	7	8	9	10
Term	$2001$	$2002$	$2003$	$2000$	$2005$	$1998$	$2007$	$1996$	$2009$	$1994$

For example, term 5 is $2002 + 2003 - 2000 = 2005$ , and term 6 is $2003 + 2000 - 2005 = 1998$ .
2. Look at the even positions only: $2002, 2000, 1998, 1996, 1994$ . Each is $2$ less than the one before. (The odd positions go up by $2$ .)
3. Why does this keep going? Rearrange the rule. It says (next term) $=$ (term 3 back) $+$ (term 2 back) $-$ (term 1 back). Move the subtracted term to the other side: the sum of two neighbors equals the sum of the two neighbors just two places earlier. So $2001 + 2002$ , $2003 + 2000$ , $2005 + 1998$ , ... all equal $4003$ , and $2002 + 2003$ , $2000 + 2005$ , $1998 + 2007$ , ... all equal $4005$ . The two sums differ by $2$ forever, which is what makes each even term $2$ less than the even term before it.
4. Count the drops. From position $2$ to position $2004$ you move $2002$ places, in jumps of $2$ , so there are $2002 \div 2 = 1001$ drops of $2$ .
5. The $2004$ th term is
$2002 - 2 \times 1001 = 2002 - 2002 = 0.$

The answer is $\boxed{\textbf{(C)}\ 0}$ .

Why this works

When a sequence rule looks strange, writing out the first ten terms almost always shows a pattern, and here it splits into two simple lists (odd places and even places). Rearranging the rule so the same kind of expression is on both sides explains why the pattern lasts. Be careful when counting jumps: the number of jumps between two positions is the distance divided by the jump size.

Alternative approach

Find a formula for the even terms and test it. The $2$ nd term is $2002 = 2004 - 2$ , the $4$ th is $2000 = 2004 - 4$ , the $6$ th is $1998 = 2004 - 6$ . So the term in an even position $p$ is $2004 - p$ . For $p = 2004$ this gives $0$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The rule is $a_{n+3} = a_n + a_{n+1} - a_{n+2}$ . Move the last term across:
$a_{n+2} + a_{n+3} = a_n + a_{n+1}.$
So the sum of two neighbors is unchanged when you slide the pair two places to the right. Starting from an odd index, every pair sums to $a_1 + a_2 = 4003$ ; starting from an even index, every pair sums to $a_2 + a_3 = 4005$ .

For any even index $2k$ , use both facts:
$a_{2k} + a_{2k+1} = 4005, \qquad a_{2k+1} + a_{2k+2} = 4003.$
Subtracting, $a_{2k+2} = a_{2k} - 2$ . The even-indexed terms form an arithmetic progression with difference $-2$ : $a_2 = 2002$ , $a_4 = 2000$ , $a_6 = 1998$ , and in general $a_{2k} = 2002 - 2(k - 1) = 2004 - 2k$ .

With $2k = 2004$ , $k = 1002$ and $a_{2004} = 2004 - 2004 = 0$ .

The answer is $\boxed{\textbf{(C)}\ 0}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)
Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic (§12.6)