Let M_{n} be the n \times n matrix with entries as follows: for 1 \leq i \leq n, m_{i, i}=10; for 1 \leq i \leq 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}{8 D_{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
for n \geq 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_{1} C_{1}-a_{2} C_{2}+a_{3} C_{3}-\cdots+(-1)^{n+1} a_{n} C_{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}.