已知an是不同正数等差数列,lga1、lga2、lga4成等差数列,又bn=1/a2n,n=1,2,3,...证明bn为等比数列
问题描述:
已知an是不同正数等差数列,lga1、lga2、lga4成等差数列,又bn=1/a2n,n=1,2,3,...
证明bn为等比数列
答
因为 lga1 lga2 lga4 成等差数列 故 2lga2= lga1+lga4 即 a2^2 =a1*a4 又 an 是不同正数的等差数列 故 a4 = a1+3d a2 =a1+d (d>0,a1>0)(a1+d)^2 = a1*(a1+3d) 故 a1d = d*d 即 a1 = d an = nd bn = 1/2nd bn-1 = 1/2...