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| − | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 15]] |
| − | Given that a sequence satisfies <math> x_0=0 </math> and <math> |x_k|=|x_{k-1}+3| </math> for all integers <math> k\ge 1, </math> find the minimum possible value of <math> |x_1+x_2+\cdots+x_{2006}|. </math>
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| − | == Solution ==
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| − | If you take the recursion formula and square both sides, you get <math>x_k^2=x_{k-1}^2+6x_{k-1}+9</math>, or <math>x_k^2-x_{k-1}^2=6x_{k-1}+9</math>. If you take the sum of this from 1 to 2007, you see that
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| − | <center><math>\begin{eqnarray*}\sum_{k=1}^{2007}x_k^2-x_{k-1}^2=\sum_{k=1}^{2007}6x_{k-1}+9\\
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| − | x_{2007}^2-x_0^2=6(x_0+x_1+x_2+\cdots +x_{2006})+9*2007\\
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| − | x_{2007}^2=6(S)+9*2007\\
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| − | S=\frac{x_{2007}^2-9*2007}{6}\end{eqnarray*}</math></center>
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| − | where <math>S=x_0+x_1+x_2+\cdots +x_{2006}=x_1+x_2+\cdots +x_{2006}</math>.
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| − | Note that <math>x_{2007}</math> is a multiple of 3, and the closest square multiple of 3 to 2*2007 is <math>(3*45)^2=9*2025</math>. Therefore the smallest value of the absolute value of <math>S</math> is
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| − | <math>|S|=\left|\frac{9*2025-9*2007}{6}\right|=\frac{9*18}{6}=\boxed{027}</math>.
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| − | == See also ==
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| − | {{AIME box|year=2006|n=II|num-b=14|after=Final Problem}}
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| − | [[Category:Intermediate Algebra Problems]]
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