Question

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

Solution:

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 $\sum_{n=1}^{\infty} \frac{1}{8D_n + 1} = \sum_{n=1}^{\infty} \frac{1}{9^{n+1}} = \frac{1/81}{1 - \frac{1}{9}} = \frac{1}{72}.$

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

2011 AIME II Problem 11

Solution:

Problem 11 in Other Years