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| − | ==Problem==
| + | #redirect [[2004 AMC 12B Problems/Problem 12]] |
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| − | 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
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| − | <math>2004^\textrm{th}</math> term in this sequence?
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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>
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| − | ==Solution==
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| − | === Solution 1 ===
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| − | 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} = 1999+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>.
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| − | === Solution 2 ===
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| − | 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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| − | Hence we get that <math>a_1+a_2 ~=~ a_3+a_4 ~=~ a_5+a_6 ~= \cdots</math> and also
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| − | <math>a_2+a_3 ~=~ a_4+a_5 ~=~ a_6+a_7 ~= \cdots</math>
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| − | From the values given in the problem statement we see that <math>a_3=a_1+2</math>.
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| − | From <math>a_1+a_2 = a_3+a_4</math> we get that <math>a_4=a_2-2</math>.
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| − | From <math>a_2+a_3 = a_4+a_5</math> we get that <math>a_5=a_3+2</math>.
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| − | Following this pattern, we get <math>a_{2004} = a_{2002} - 2 = a_{2000} - 4 = \cdots = a_2 - 2002 = 0</math>.
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| − | == See also ==
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| − | {{AMC10 box|year=2004|ab=B|num-b=|num-a=}}
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