Question

设 $u_0 = \dfrac{1}{4}$ ，且对 $k \ge 0$ ，由递推式定义 $u_{k+1}$ 。该数列趋于极限 $L$ 。满足的最小 $k$ 是多少？ $u_{k+1} = 2u_k - 2u_k^2.$ $|u_k - L| \le \dfrac{1}{2^{1000}}?$

Set $u_0 = \dfrac{1}{4},$ and for $k \ge 0$ let $u_{k+1}$ be determined by the recurrence $u_{k+1} = 2u_k - 2u_k^2.$ This sequence tends to a limit; call it $L.$ What is the least value of $k$ such that $|u_k - L| \le \dfrac{1}{2^{1000}}?$

$10$

$87$

$123$

$329$

$401$

答案：A

知识点：递推换元法

难度评级：2230

解答：

极限满足 $L = 2L - 2L^2$ ，得 $L = \tfrac12$ 。令 $v_k = 1 - 2u_k$ 。于是 $\begin{aligned} v_{k+1} &= 1 - 2u_{k+1} \\ &= 1 - 4u_k + 4u_k^2 \\ &= (1 - 2u_k)^2 = v_k^2. \end{aligned}$

由 $v_0 = 1 - 2 \cdot \tfrac14 = \tfrac12$ ，可得 $v_k = \left(\tfrac12\right)^{2^k}$ ，所以 $|u_k - L| = \dfrac{|v_k|}{2} = 2^{-2^k - 1}$ 。

需要 $2^k + 1 \ge 1000$ ，也就是 $2^k \ge 999$ 。满足条件的最小 $k$ 是 $10$ ，因为 $2^{10} = 1024$ 。

所以正确答案是 A。

The limit satisfies $L = 2L - 2L^2,$ giving $L = \tfrac12.$ Let $v_k = 1 - 2u_k.$ Then $\begin{aligned} v_{k+1} &= 1 - 2u_{k+1} \\ &= 1 - 4u_k + 4u_k^2 \\ &= (1 - 2u_k)^2 = v_k^2. \end{aligned}$

Since $v_0 = 1 - 2 \cdot \tfrac14 = \tfrac12,$ we get $v_k = \left(\tfrac12\right)^{2^k},$ so $|u_k - L| = \dfrac{|v_k|}{2} = 2^{-2^k - 1}.$

We need $2^k + 1 \ge 1000,$ i.e. $2^k \ge 999.$ The least such $k$ is $10,$ since $2^{10} = 1024.$

Thus, the correct answer is A.

2021 AMC 12B Fall 第 18 题

答案：A

难度评级：2230

解答：

其他年份的第 18 题