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AIME 2011b Problem 11

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Let MnM_n be the n×nn \times n matrix with entries as follows: for 1≤i≤n1 \le i \le n, mi,i=10m_{i,i} = 10; for 1≤i≤n−11 \le i \le n - 1, mi+1,i=mi,i+1=3m_{i+1,i} = m_{i,i+1} = 3; all other entries in MnM_n are zero. Let DnD_n be the determinant of matrix MnM_n. Then ∑n=1∞18Dn+1\sum_{n=1}^{\infty} \frac{1}{8D_n+1} can be represented as pq\frac{p}{q}, where pp and qq are relatively prime positive integers. Find p+qp + q. Note: The determinant of the 1×11 \times 1 matrix [a][a] is aa, and the determinant of the 2×22 \times 2 matrix [abcd]=ad−bc\left[ {\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right] = ad - bc; for n≥2n \ge 2, the determinant of an n×nn \times n matrix with first row or first column a1a_1 a2a_2 a3a_3 …\dots ana_n is equal to a1C1−a2C2+a3C3−⋯+(−1)n+1anCna_1C_1 - a_2C_2 + a_3C_3 - \dots + (-1)^{n+1}a_nC_n, where CiC_i is the determinant of the (n−1)×(n−1)(n - 1) \times (n - 1) matrix formed by eliminating the row and column containing aia_i.

Dataset text AIME 1983 to 2026. Quoted with source.

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