Question

数列 $(a_n)$ 满足 $a_1 = 1$ ，且对 $n \ge 1$ 有 $5^{(a_{n+1} - a_n)} - 1 = \frac{1}{n + \frac{2}{3}}$ 。令 $k$ 为大于 $1$ 的最小整数，使得 $a_k$ 是整数。求 $k$ 。

The sequence $(a_n)$ satisfies $a_1 = 1$ and $5^{(a_{n+1} - a_n)} - 1 = \frac{1}{n + \frac{2}{3}}$ for $n \ge 1.$ Let $k$ be the least integer greater than $1$ for which $a_k$ is an integer. Find $k.$

答案

答案：41

知识点：裂项相消对数模运算

难度评级：2450

解答：

该关系给出 $5^{a_{n+1} - a_n} = 1 + \frac{3}{3n + 2} = \frac{3n+5}{3n+2}$ 。将 $n = 1, 2, \ldots, k - 1$ 的这些等式相乘，发生望远镜相消： $5^{a_k - a_1} = \frac{3k + 2}{5},$ 所以 $\begin{aligned} a_k &= 1 + \log_5 \frac{3k+2}{5} \\ &= \log_5 (3k + 2). \end{aligned}$

因此 $a_k$ 为整数当且仅当 $3k + 2$ 是 $5$ 的幂。由于 $5^j \equiv 2^j \pmod 3$ ，只有奇数指数 $j$ 才能给出形如 $3k + 2$ 的数。幂 $5^1 = 5$ 给出 $k = 1$ 被排除；下一个 $5^3 = 125 = 3 \cdot 41 + 2$ ，给出 $k = 41$ 。

The relation says $5^{a_{n+1} - a_n} = 1 + \frac{3}{3n + 2} = \frac{3n+5}{3n+2}.$ Multiplying these equations for $n = 1, 2, \ldots, k - 1$ telescopes: $5^{a_k - a_1} = \frac{3k + 2}{5},$ so $\begin{aligned} a_k &= 1 + \log_5 \frac{3k+2}{5} \\ &= \log_5 (3k + 2). \end{aligned}$

Thus $a_k$ is an integer exactly when $3k + 2$ is a power of $5.$ Since $5^j \equiv 2^j \pmod 3,$ only odd exponents $j$ give numbers of the form $3k + 2.$ The power $5^1 = 5$ gives $k = 1,$ which is excluded, and the next, $5^3 = 125 = 3 \cdot 41 + 2,$ gives $k = 41.$

2009 AIME I 第 7 题

答案：41

难度评级：2450

解答：

其他年份的第 7 题