等差数列{an}中an=9n-8.对任意m∈N﹡,将数列{an}中落入区间（9^m,9^2m）内的项的个数记为bm,求数列{bm}

Question

等差数列{an} 中
an=9n-8.对任意m ∈N﹡,将数列{an} 中落入区间（9^m ,9^2m ）内的项的个数记为bm ,求数列{bm}

Answer 1

9*9^(m-1) - 8 = a[9^(m-1)] = 9^m - 8 < 9^m < 9^m + 1 = 9*[9^(m-1) + 1] - 8 = a[9^(m-1) + 1].9*9^(2m-1) - 8 = a[9^(2m-1)] = 9^(2m)-8 < 9^(2m) < 9^(2m) + 1 = 9*[9^(2m-1)+1] - 8 = a[9^(2m-1) +1].a(n)中落...