设数列【An】的前n项和为Sn,A1=10,An+1=9Sn+10.设Bn=lgAn,求证数列【Bn】为等差数列
问题描述：
设数列【An】的前n项和为Sn,A1=10,An+1=9Sn+10.设Bn=lgAn,求证数列【Bn】为等差数列
答
An+1=9Sn+10
An=9S(n-1)+10
An=Sn-S(n-1)=(1/9)[A(n+1)-An]
A(n+1)/An=10
所以为等比数列 A1=10,q=10
An=10*10^(n-1)=10^n
Bn=lgAn=n
B(n+1)-Bn=n-(n-1)=1
1为常数,所以是等差数列