Question

递增的等比数列 $x_0, x_1, x_2, \ldots$ 完全由 $3$ 的整数次幂组成。已知 $\sum_{n=0}^{7} \log_3(x_n) = 308$ 且 $56 \le \log_3\left(\sum_{n=0}^{7} x_n\right) \le 57,$ 求 $\log_3(x_{14})$ 。

The increasing geometric sequence $x_0, x_1, x_2, \ldots$ consists entirely of integral powers of $3.$ Given that $\sum_{n=0}^{7} \log_3(x_n) = 308$ and $56 \le \log_3\left(\sum_{n=0}^{7} x_n\right) \le 57,$ find $\log_3(x_{14}).$

答案

答案：91

知识点：等比数列对数极限情形界定

难度评级：2650

解答：

每一项都是 $3$ 的幂，公比也是两个 $3$ 的幂之商，所以 $x_n = 3^{a + bn}$ ，其中 $a$ 和 $b$ 为整数；又因为数列递增， $b \ge 1$ 。第一个条件给出 $\sum_{n=0}^{7} (a + bn) = 8a + 28b = 308,$ 即 $2a + 7b = 77.$

对第二个条件， $x_7 = 3^{a+7b}$ 是最大项，并且 $\begin{aligned} x_7 &\lt \sum_{n=0}^{7} x_n \\ &\lt x_7\left(1 + \tfrac{1}{3} + \tfrac{1}{9} + \cdots\right) \\ &= \tfrac{3}{2}\,x_7 \lt 3x_7, \end{aligned}$ 所以 $\log_3\left(\sum x_n\right)$ 严格介于 $a + 7b$ 和 $a + 7b + 1$ 之间。给定界限迫使 $a + 7b = 56$ 。

从 $2a + 7b = 77$ 中减去它，得 $a = 21$ ，进而 $b = 5$ 。因此 $\log_3(x_{14}) = a + 14b = 21 + 70$ $= 91$ 。

Every term is a power of $3$ and the ratio is a quotient of powers of $3,$ so $x_n = 3^{a + bn}$ for integers $a$ and $b,$ with $b \ge 1$ since the sequence increases. The first condition gives $\sum_{n=0}^{7} (a + bn) = 8a + 28b = 308,$ i.e. $2a + 7b = 77.$

For the second condition, $x_7 = 3^{a+7b}$ is the largest term, and $\begin{aligned} x_7 &\lt \sum_{n=0}^{7} x_n \\ &\lt x_7\left(1 + \tfrac{1}{3} + \tfrac{1}{9} + \cdots\right) \\ &= \tfrac{3}{2}\,x_7 \lt 3x_7, \end{aligned}$ so $\log_3\left(\sum x_n\right)$ lies strictly between $a + 7b$ and $a + 7b + 1.$ The given bounds then force $a + 7b = 56.$

Subtracting from $2a + 7b = 77$ yields $a = 21,$ then $b = 5.$ Therefore $\log_3(x_{14}) = a + 14b = 21 + 70$ $= 91.$

2007 AIME II 第 12 题

答案：91

难度评级：2650

解答：

其他年份的第 12 题