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2004 AMC 12B Problems/Problem 12: Difference between revisions

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<math> \mathrm{(A) \ } -2004 \qquad \mathrm{(B) \ } -2 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } 4003 \qquad \mathrm{(E) \ } 6007 </math>
<math> \mathrm{(A) \ } -2004 \qquad \mathrm{(B) \ } -2 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } 4003 \qquad \mathrm{(E) \ } 6007 </math>


=== Solution 1 ===
== Solution 1 ==


We already know that <math>a_1=2001</math>, <math>a_2=2002</math>, <math>a_3=2003</math>, and <math>a_4=2000</math>. Let's compute the next few terms to get the idea how the sequence behaves. We get <math>a_5 = 2002+2003-2000 = 2005</math>, <math>a_6=2003+2000-2005=1998</math>, <math>a_7=2000+2005-1998=2007</math>, and so on.
We already know that <math>a_1=2001</math>, <math>a_2=2002</math>, <math>a_3=2003</math>, and <math>a_4=2000</math>. Let's compute the next few terms to get the idea how the sequence behaves. We get <math>a_5 = 2002+2003-2000 = 2005</math>, <math>a_6=2003+2000-2005=1998</math>, <math>a_7=2000+2005-1998=2007</math>, and so on.
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We can now discover the following pattern: <math>a_{2k+1} = 2001+2k</math> and <math>a_{2k}=2004-2k</math>. This is easily proved by induction. It follows that <math>a_{2004}=a_{2\cdot 1002} = 2004 - 2\cdot 1002 = \boxed{0}</math>.
We can now discover the following pattern: <math>a_{2k+1} = 2001+2k</math> and <math>a_{2k}=2004-2k</math>. This is easily proved by induction. It follows that <math>a_{2004}=a_{2\cdot 1002} = 2004 - 2\cdot 1002 = \boxed{0}</math>.


=== Solution 2 ===
== Solution 1 but in a more not smart way==
 
Subtract 2000 from each of the terms so the sequence turns into 1, 2, 3, 0, 5, -2, 7, -4, 9, -6... (1+2-3=0, 2+3-0=5, 3+0-5=-2, 0+5-(-2)=7, 5+(-2)-7=-4, etc.). Quickly notice that after taking away 1 and 2, the 2nd term is 0, the 4th term is -2, the 6th term is -4, the 8th term is -6, etc. Thus, in the sequence without 1 and 2, the (2n)th term has a value of -2(n-1). Therefore, to find the 2004th term in the original sequence, we take away the first 2 values to form the new sequence so the value of the 2004th term in the original sequence is the 2002nd term in the new sequence. 2002 is 2 times 1001. Thus, the value is -2(1001-1)= -2000. BUT, REMEMBER THAT AT THE START WE TOOK AWAY 2000 FROM EACH TERM. ADD -2000 TO 2000 TO GET <math>\boxed{0}</math>. LOLOL
 
== Solution 2 ==


Note that the recurrence <math>a_n+a_{n+1}-a_{n+2}~=~a_{n+3}</math> can be rewritten as <math>a_n+a_{n+1} ~=~ a_{n+2}+a_{n+3}</math>.  
Note that the recurrence <math>a_n+a_{n+1}-a_{n+2}~=~a_{n+3}</math> can be rewritten as <math>a_n+a_{n+1} ~=~ a_{n+2}+a_{n+3}</math>.  
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Following this pattern, we get <math>a_{2004} = a_{2002} - 2 = a_{2000} - 4 = \cdots = a_2 - 2002 = \boxed{0}</math>.
Following this pattern, we get <math>a_{2004} = a_{2002} - 2 = a_{2000} - 4 = \cdots = a_2 - 2002 = \boxed{0}</math>.


=== Solution 3 ===
== Solution 3 ==


Our recurrence is <math>a_n+a_{n+1}-a_{n+2}~=~a_{n+3}</math>, so we get <math>r^3+r^2-r-1 = 1</math>, so <math>(r-1)(r+1)^2 = 1</math>, so our formula for the recurrence is <math>a_n = A + (B + Cn)(-1)^n</math>.
Our recurrence is <math>a_n+a_{n+1}-a_{n+2}~=~a_{n+3}</math>, so we get <math>r^3+r^2-r-1 = 1</math>, so <math>(r-1)(r+1)^2 = 1</math>, so our formula for the recurrence is <math>a_n = A + (B + Cn)(-1)^n</math>.
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So, <math>a_{2004} = 2002 - 2002 = 0.</math>
So, <math>a_{2004} = 2002 - 2002 = 0.</math>
~ ilovepizza2020


== See also ==
== See also ==

Latest revision as of 11:03, 15 December 2024

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 1 but in a more not smart way

Subtract 2000 from each of the terms so the sequence turns into 1, 2, 3, 0, 5, -2, 7, -4, 9, -6... (1+2-3=0, 2+3-0=5, 3+0-5=-2, 0+5-(-2)=7, 5+(-2)-7=-4, etc.). Quickly notice that after taking away 1 and 2, the 2nd term is 0, the 4th term is -2, the 6th term is -4, the 8th term is -6, etc. Thus, in the sequence without 1 and 2, the (2n)th term has a value of -2(n-1). Therefore, to find the 2004th term in the original sequence, we take away the first 2 values to form the new sequence so the value of the 2004th term in the original sequence is the 2002nd term in the new sequence. 2002 is 2 times 1001. Thus, the value is -2(1001-1)= -2000. BUT, REMEMBER THAT AT THE START WE TOOK AWAY 2000 FROM EACH TERM. ADD -2000 TO 2000 TO GET $\boxed{0}$. LOLOL

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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