Problem:
Let Mn​ be the n×n matrix with entries as follows: for 1≤i≤n, mi,i​=10; for 1≤i≤n−1,mi+1,i​=mi,i+1​=3; all other entries in Mn​ are zero. Let Dn​ be the determinant of matrix Mn​. Then ∑n=1∞​8Dn​+11​ can be represented as qp​, where p and q are relatively prime positive integers. Find p+q.
Note: The determinant of the 1×1 matrix [a] is a, and the determinant of the 2×2 matrix
[ac​bd​]=ad−bc;
for n≥2, the determinant of an n×n matrix with first row or first column a1​a2​a3​…an​ is equal to a1​C1​−a2​C2​+a3​C3​−⋯+(−1)n+1an​Cn​, where Ci​ is the determinant of the (n−1)×(n−1) matrix formed by eliminating the row and column containing ai​.
Solution:
Let Dn​= the determinant of Mn​. The cofactor expansion of Dn​ along the first column of Mn​ and then along the first row of the matrix formed by omitting the second row and first column of Mn​ shows that Dn​= 10Dn−1​−9Dn−2​. This is a second-order recurrence relation. It can be shown that its solution is a linear combination of expressions of the form kn. Then the equation simplifies to k2−10k+9=0, which has solutions 9 and 1. Thus Dn​=a⋅9n+b⋅1n, and substituting the initial conditions D1​=10 and D2​=10⋅10−3⋅3=91 yields Dn​=89​⋅9n−81​⋅1n= 89n+1−1​=8(9−1)(9n+9n−1+⋯+9+1)​=9n+9n−1+⋯+9+1. The requested sum is thus ∑n=1∞​9n+11​=1−91​811​​=721​, and p+q=73​.
The problems on this page are the property of the MAA's American Mathematics Competitions