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
[acbd]=ad−bc;
for n≥2, the determinant of an n×n matrix with first row or first column a1a2a3…an is equal to a1C1−a2C2+a3C3−⋯+(−1)n+1anCn, 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=1−91811=721, and p+q=73.
The problems on this page are the property of the MAA's American Mathematics Competitions