2006 AIME A Problems/Problem 15

Problem

Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1,$ find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|.$

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Solution

If you take the recursion formula and square both sides, you get $x_k^2=x_{k-1}^2+6x_{k-1}+9$ , or $x_k^2-x_{k-1}^2=6x_{k-1}+9$ . If you take the sum of this from 1 to 2007, you see that

$\begin{eqnarray*}\sum_{k=1}^{2007}x_k^2-x_{k-1}^2=\sum_{k=1}^{2007}6x_{k-1}+9\\

x_{2007}^2-x_0^2=6(x_0+x_1+x_2+\cdots +x_{2006})+9*2007\\ x_{2007}^2=6(S)+9*2007\\

S=\frac{x_{2007}^2-9*2007}{6}\end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)

where $S=x_0+x_1+x_2+\cdots +x_{2006}=x_1+x_2+\cdots +x_{2006}$ .

Note that $x_{2007}$ is a multiple of 3, and the closest square multiple of 3 to 2*2007 is $(3*45)^2=9*2025$ . Therefore the smallest value of the absolute value of $S$ is

$|S|=\left|\frac{9*2025-9*2007}{6}\right|=\frac{9*18}{6}=\boxed{027}$ .

2006 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Final Problem
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2006 AIME A Problems/Problem 15

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