2011 AIME II Problem 11

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 \left[ {\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right] = 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 - \ldots + (-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.

Leading zeroes must be inputted, so if your answer is 34, then input 034

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.