Question

设 $M_n$ 为如下 $n \times n$ 矩阵：对于 $1 \le i \le n$ ， $m_{i,i} = 10$ ；对于 $1 \le i \le n - 1$ ， $m_{i+1,i} = m_{i,i+1} = 3$ ； $M_n$ 中所有其他元素都是零。设 $D_n$ 为矩阵 $M_n$ 的行列式。那么 $\sum_{n=1}^{\infty} \frac{1}{8D_n + 1}$ 可以表示为 $\frac{p}{q}$ ，其中 $p$ 和 $q$ 是互质的正整数。求 $p + q$ 。

注： $1 \times 1$ 矩阵 $[a]$ 的行列式为 $a$ ，而 $2 \times 2$ 矩阵的行列式为 $\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc;$ 对 $n \ge 2,$ 一个 $n \times n$ 矩阵沿第一行或第一列 $a_1$ $a_2$ $a_3$ $\ldots$ $a_n$ 展开的行列式等于 $a_1C_1 - a_2C_2$ $+ a_3C_3 - \cdots$ $+ (-1)^{n+1}a_nC_n$ ，其中 $C_i$ 是删去包含 $a_i$ 的行和列后形成的 $(n - 1) \times (n - 1)$ 矩阵的行列式。

Let $M_n$ be the $n \times n$ matrix with entries as follows: for $1 \le i \le n,$ $m_{i,i} = 10;$ for $1 \le i \le n - 1,$ $m_{i+1,i} = m_{i,i+1} = 3;$ all other entries in $M_n$ are zero. Let $D_n$ be the determinant of matrix $M_n.$ Then $\sum_{n=1}^{\infty} \frac{1}{8D_n + 1}$ can be represented as $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

Note: The determinant of the $1 \times 1$ matrix $[a]$ is $a,$ and the determinant of the $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc;$ for $n \ge 2,$ the determinant of an $n \times n$ matrix with first row or first column $a_1$ $a_2$ $a_3$ $\ldots$ $a_n$ is equal to $a_1C_1 - a_2C_2$ $+ a_3C_3 - \cdots$ $+ (-1)^{n+1}a_nC_n,$ where $C_i$ is the determinant of the $(n - 1) \times (n - 1)$ matrix formed by eliminating the row and column containing $a_i.$

答案

答案：73

知识点：行列式递推等比数列

难度评级：2920

解答：

沿第一行展开 $D_n$ ，得到 $10 D_{n-1}$ 减去 $3$ 乘以一个余子式；该余子式的第一列为 $(3, 0, \ldots, 0)$ ；再沿其第一列展开会留下 $3 D_{n-2}$ 。因此 $D_n = 10 D_{n-1} - 9 D_{n-2},$ 且 $D_1 = 10$ ， $D_2 = 100 - 9 = 91$ 。

特征方程 $k^2 = 10k - 9$ 的根为 $9$ 和 $1$ ，代入初值可得 $D_n = \frac{9^{n+1} - 1}{8}$ 。因此 $8D_n + 1 = 9^{n+1}$ ，并且 $\begin{aligned} \sum_{n=1}^{\infty} \frac{1}{8D_n + 1} &= \sum_{n=1}^{\infty} \frac{1}{9^{n+1}} \\ &= \frac{1/81}{1 - \frac{1}{9}} = \frac{1}{72}. \end{aligned}$

所以 $\frac{p}{q} = \frac{1}{72}$ ，且 $p + q = 73$ 。

Expanding $D_n$ along the first row gives $10 D_{n-1}$ minus $3$ times a cofactor whose first column is $(3, 0, \ldots, 0);$ expanding that cofactor down its first column leaves $3 D_{n-2}.$ Hence $D_n = 10 D_{n-1} - 9 D_{n-2},$ with $D_1 = 10$ and $D_2 = 100 - 9 = 91.$

The characteristic equation $k^2 = 10k - 9$ has roots $9$ and $1,$ and fitting the initial values gives $D_n = \frac{9^{n+1} - 1}{8}.$ Therefore $8D_n + 1 = 9^{n+1},$ and $\begin{aligned} \sum_{n=1}^{\infty} \frac{1}{8D_n + 1} &= \sum_{n=1}^{\infty} \frac{1}{9^{n+1}} \\ &= \frac{1/81}{1 - \frac{1}{9}} = \frac{1}{72}. \end{aligned}$

Thus $\frac{p}{q} = \frac{1}{72}$ and $p + q = 73.$

2011 AIME II 第 11 题

答案：73

难度评级：2920

解答：

其他年份的第 11 题