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

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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】为等差数列设数列【An】的前n项和为Sn,A1=10,An+1=9Sn+10.设Bn=lgAn,求

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
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

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
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

这个是不是有问题啊 Sn为An的前n项合
那么S1=A1
根据An+1=9Sn+10 那么A1+1=9S1+10 因为S1=A1
所以A1+1=9A1+10 得出A1=-9/8 和给出的A1=10不相符啊

因此 bn 是等差数列 b8 = a8/8^(8-8) = 8/8 = 8 bn = n -------------------- an/8^(n-8) = n 所以 an = n * 8^(n-8) ------------------------- Sn = a8 + a8 + a8 + …… + a<n-8> + an = 8 + 8*8 +

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