Question

令 $f(x) = \left(x^2 + 3x + 2\right)^{\cos(\pi x)}$ 。求所有满足 $\left|\sum_{k=1}^{n} \log_{10} f(k)\right| = 1.$ 的正整数 $n$ 之和。

Let $f(x) = \left(x^2 + 3x + 2\right)^{\cos(\pi x)}.$ Find the sum of all positive integers $n$ for which $\left|\sum_{k=1}^{n} \log_{10} f(k)\right| = 1.$

答案

答案：21

知识点：裂项相消对数奇偶性

难度评级：2450

解答：

因为 $\cos(\pi k) = (-1)^k$ ，且 $k^2 + 3k + 2 = (k+1)(k+2)$ ，所以 $f(k) = \left[(k+1)(k+2)\right]^{(-1)^k}$ 。这些对数之和等于乘积 $\prod_{k=1}^n f(k)$ 的对数，而乘积会望远镜消去：相邻因子 $\frac{1}{(k+1)(k+2)}$ 和 $(k+2)(k+3)$ 只留下边界项。

当 $n$ 为偶数时，乘积为 $\frac{3 \cdot 4 \cdots (n+2)}{2 \cdot 3 \cdots (n+1)} = \frac{n+2}{2}$ ，当 $n$ 为奇数时，乘积为 $\frac{3 \cdot 4 \cdots (n+1)}{2 \cdot 3 \cdots (n+2)} = \frac{1}{2(n+2)}$ 。对数的绝对值等于 $1$ ，当且仅当乘积为 $10$ 或 $\frac{1}{10}$ 。

偶数 $n$ 时， $\frac{n+2}{2} = 10$ 给出 $n = 18$ ；奇数 $n$ 时， $2(n+2) = 10$ 给出 $n = 3$ 。所求和为 $18 + 3 = 21$ 。

Since $\cos(\pi k) = (-1)^k$ and $k^2 + 3k + 2 = (k+1)(k+2),$ we have $f(k) = \left[(k+1)(k+2)\right]^{(-1)^k}.$ The sum of the logarithms is the log of the product $\prod_{k=1}^n f(k),$ which telescopes: consecutive factors $\frac{1}{(k+1)(k+2)}$ and $(k+2)(k+3)$ leave only boundary terms.

For even $n$ the product is $\frac{3 \cdot 4 \cdots (n+2)}{2 \cdot 3 \cdots (n+1)} = \frac{n+2}{2},$ and for odd $n$ it is $\frac{3 \cdot 4 \cdots (n+1)}{2 \cdot 3 \cdots (n+2)} = \frac{1}{2(n+2)}.$ The absolute value of the log equals $1$ exactly when the product is $10$ or $\frac{1}{10}.$

For even $n,$ $\frac{n+2}{2} = 10$ gives $n = 18;$ for odd $n,$ $2(n+2) = 10$ gives $n = 3.$ The requested sum is $18 + 3 = 21.$

2014 AIME II 第 7 题

答案：21

难度评级：2450

解答：

其他年份的第 7 题