Question

已知一个数列满足 $x_0 = 0$ ，且对所有整数 $k \ge 1$ 都有 $|x_k| = |x_{k-1} + 3|$ 。求 $|x_1 + x_2 + \cdots + x_{2006}|$ 的最小可能值。

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 $|x_1 + x_2 + \cdots + x_{2006}|.$

答案

答案：27

知识点：递推裂项相消奇偶性最优化

难度评级：3370

解答：

将递推式平方得 $x_k^2 = (x_{k-1} + 3)^2$ $= x_{k-1}^2 + 6 x_{k-1} + 9$ 。对 $k = 1$ 到 $2007$ 求和并望远镜相消： $x_{2007}^2 = x_0^2 + 6 \sum_{k=0}^{2006} x_k + 9 \cdot 2007,$ 所以由 $x_0 = 0$ ， $\begin{aligned} &x_1 + x_2 + \cdots + x_{2006} \\ &= \frac{x_{2007}^2 - 18063}{6}. \end{aligned}$

归纳可知每个 $x_k$ 都是 $3$ 的倍数，且奇偶性与 $k$ 相同，所以 $x_{2007}$ 是 $3$ 的奇数倍。要使 $\left|x_{2007}^2 - 18063\right|$ 最小，取平方最接近 $18063$ 的 $3$ 的奇数倍，即 $\pm 135$ 。其中 $135^2 = 18225$ ，得到 $\frac{18225 - 18063}{6} = 27$ （相邻的 $129$ 和 $141$ 分别给出 $237$ 和 $303$ ）。

这个值可以达到：当 $k \le 45$ 时取 $x_k = 3k$ ；此后，偶数 $k$ 时令 $x_k = -138$ ，奇数 $k$ 时令 $x_k = 135$ 。于是 $x_{2007} = 135$ ，和为 $27$ 。所以最小值为 $27$ 。

Squaring the recurrence gives $x_k^2 = (x_{k-1} + 3)^2$ $= x_{k-1}^2 + 6 x_{k-1} + 9.$ Summing for $k = 1$ to $2007$ telescopes: $x_{2007}^2 = x_0^2 + 6 \sum_{k=0}^{2006} x_k + 9 \cdot 2007,$ so with $x_0 = 0,$ $\begin{aligned} &x_1 + x_2 + \cdots + x_{2006} \\ &= \frac{x_{2007}^2 - 18063}{6}. \end{aligned}$

Induction shows each $x_k$ is a multiple of $3$ whose parity matches that of $k,$ so $x_{2007}$ is an odd multiple of $3.$ To minimize $\left|x_{2007}^2 - 18063\right|,$ take the odd multiple of $3$ whose square is nearest $18063:$ that is $\pm 135,$ with $135^2 = 18225,$ giving $\frac{18225 - 18063}{6} = 27$ (the neighbors $129$ and $141$ give $237$ and $303$ ).

This value is attained: take $x_k = 3k$ for $k \le 45,$ and thereafter alternate $x_k = -138$ for even $k$ and $x_k = 135$ for odd $k;$ then $x_{2007} = 135$ and the sum is $27.$ So the minimum is $27.$

2006 AIME I 第 15 题

答案：27

难度评级：3370

解答：

其他年份的第 15 题