2006 AIME A Problems/Problem 15

Problem

Given that a sequence satisfies and for all integers find the minimum possible value of

Solution

If you take the recursion formula and square both sides, you get , or . 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 .

Note that x_{2007} is a multiple of 3, and the closest square multiple of 3 to 2*2007 is . Therefore the smallest value of the absolute value of is

.