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Difference between revisions of "2004 AMC 12B Problems/Problem 12"

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{{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #12]] and [[2004 AMC 10B Problems|2004 AMC 10B #19]]}}
 
{{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #12]] and [[2004 AMC 10B Problems|2004 AMC 10B #19]]}}
  
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==Problem==
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===Problem===
  
 
In the sequence <math>2001</math>, <math>2002</math>, <math>2003</math>, <math>\ldots</math> , 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 <math>2001 + 2002 - 2003 = 2000</math>. What is the
 
In the sequence <math>2001</math>, <math>2002</math>, <math>2003</math>, <math>\ldots</math> , 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 <math>2001 + 2002 - 2003 = 2000</math>. What is the

Revision as of 10:40, 13 October 2021

The following problem is from both the 2004 AMC 12B #12 and 2004 AMC 10B #19, so both problems redirect to this page.

Problem

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?

$\mathrm{(A) \ } -2004 \qquad \mathrm{(B) \ } -2 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } 4003 \qquad \mathrm{(E) \ } 6007$

Solution 1

We already know that $a_1=2001$, $a_2=2002$, $a_3=2003$, and $a_4=2000$. Let's compute the next few terms to get the idea how the sequence behaves. We get $a_5 = 2002+2003-2000 = 2005$, $a_6=2003+2000-2005=1998$, $a_7=2000+2005-1998=2007$, and so on.

We can now discover the following pattern: $a_{2k+1} = 2001+2k$ and $a_{2k}=2004-2k$. This is easily proved by induction. It follows that $a_{2004}=a_{2\cdot 1002} = 2004 - 2\cdot 1002 = \boxed{0}$.

Solution 2

Note that the recurrence $a_n+a_{n+1}-a_{n+2}~=~a_{n+3}$ can be rewritten as $a_n+a_{n+1} ~=~ a_{n+2}+a_{n+3}$.

Hence we get that $a_1+a_2 ~=~ a_3+a_4 ~=~ a_5+a_6 ~= \cdots$ and also $a_2+a_3 ~=~ a_4+a_5 ~=~ a_6+a_7 ~= \cdots$

From the values given in the problem statement we see that $a_3=a_1+2$.

From $a_1+a_2 = a_3+a_4$ we get that $a_4=a_2-2$.

From $a_2+a_3 = a_4+a_5$ we get that $a_5=a_3+2$.

Following this pattern, we get $a_{2004} = a_{2002} - 2 = a_{2000} - 4 = \cdots = a_2 - 2002 = \boxed{0}$.

Solution 3

Our recurrence is $a_n+a_{n+1}-a_{n+2}~=~a_{n+3}$, so we get $r^3+r^2-r-1 = 1$, so $(r-1)(r+1)^2 = 1$, so our formula for the recurrence is $a_n = A + (B + Cn)(-1)^n$.

Substituting our starting values gives us $a_n = 2002 + (2 - n)(-1)^n$.

So, $a_{2004} = 2002 - 2002 = 0.$

~ ilovepizza2020

See also

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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