Question

设 $\{a_k\}_{k=1}^{2011}$ 为如下定义的实数序列： $a_1 = 0.201$ ， $a_2 = (0.2011)^{a_1}$ ， $a_3 = (0.20101)^{a_2}$ ， $a_4 = (0.201011)^{a_3}$ 。更一般地， $a_k = \begin{cases} \tiny \left(0.\underbrace{20101\ldots0101}_{k+2 \text{ digits}}\right)^{a_{k-1}}, & \tiny \text{if } k \text{ is odd,} \\ \tiny \left(0.\underbrace{20101\ldots01011}_{k+2 \text{ digits}}\right)^{a_{k-1}}, & \tiny \text{if } k \text{ is even.} \end{cases}$

将序列 $\{a_k\}_{k=1}^{2011}$ 中的数按递减顺序重新排列，得到新序列 $\{b_k\}_{k=1}^{2011}$ 。求所有满足 $a_k = b_k$ 的整数 $k$ （其中 $1 \le k \le 2011$ ）的和。

Let $\{a_k\}_{k=1}^{2011}$ be the sequence of real numbers defined by $a_1 = 0.201,$ $a_2 = (0.2011)^{a_1},$ $a_3 = (0.20101)^{a_2},$ and $a_4 = (0.201011)^{a_3},$ and more generally $a_k = \begin{cases} \tiny \left(0.\underbrace{20101\ldots0101}_{k+2 \text{ digits}}\right)^{a_{k-1}}, & \tiny \text{if } k \text{ is odd,} \\ \tiny \left(0.\underbrace{20101\ldots01011}_{k+2 \text{ digits}}\right)^{a_{k-1}}, & \tiny \text{if } k \text{ is even.} \end{cases}$

Rearranging the numbers in the sequence $\{a_k\}_{k=1}^{2011}$ in decreasing order produces a new sequence $\{b_k\}_{k=1}^{2011}.$ What is the sum of all the integers $k,$ $1 \le k \le 2011,$ such that $a_k = b_k?$

$671$

$1006$

$1341$

$2011$

$2012$

答案：C

知识点：指数不等式

难度评级：2460

解答：

因为每个底数都严格介于 $0$ 和 $1$ 之间，函数 $t \mapsto (\text{base})^t$ 是递减的，而对 $b \gt 0$ ， $t \mapsto t^b$ 是递增的。比较各项可得序列的大小顺序为 $\begin{aligned} &1 \gt a_2 \gt a_4 \gt \cdots \gt a_{2010} \\ &\gt a_{2011} \gt a_{2009} \\ &\gt \cdots \gt a_1 \gt 0. \end{aligned}$

因此在递减排列中，偶数下标项先出现，然后是奇数下标项按反向出现。某项满足 $a_k = b_k$ 当且仅当它的位置等于它的下标；对递减排列的奇数尾部，这要求 $2(k - 1006) = 2011 - k$ 。

解得 $3k = 4023$ ，所以 $k = 1341$ ，这是唯一固定下标，因此所求和为 $1341$ 。

因此，正确答案是 C。

Because each base lies strictly between $0$ and $1,$ the function $t \mapsto (\text{base})^t$ is decreasing, while $t \mapsto t^b$ is increasing for $b \gt 0.$ Comparing terms shows the sequence orders as $\begin{aligned} &1 \gt a_2 \gt a_4 \gt \cdots \gt a_{2010} \\ &\gt a_{2011} \gt a_{2009} \\ &\gt \cdots \gt a_1 \gt 0. \end{aligned}$

So in the decreasing arrangement, the even-indexed terms come first, then the odd-indexed terms in reverse. A term satisfies $a_k = b_k$ exactly when its position equals its index, which for the descending odd tail requires $2(k - 1006) = 2011 - k.$

Solving gives $3k = 4023,$ so $k = 1341,$ the unique fixed index, and the sum is $1341.$

Thus, the correct answer is C.

2012 AMC 12A 第 24 题

答案：C

难度评级：2460

解答：

其他年份的第 24 题