∴Sn+1=n+1-5an+1-85,n∈N*.
两式作差得an+1=1-5an+1+5an,即6(an+1-1)=5(an-1),即(an+1-1)=
5 |
6 |
故{an-1}是等比数列
(2)由(1)Sn+1=n+1-5an+1-85,n∈N*.得Sn+1=n+1-5(Sn+1-Sn)-85,n∈N*.
得6Sn+1=n+5Sn-84,即6[Sn+1-(n+1)]=5(Sn-n)-90,
即Sn+1-(n+1)=
5 |
6 |
整理得Sn+1-(n+1)+90=
5 |
6 |
故{Sn-n+90}是一个等比数列,其公比为
5 |
6 |
故{Sn-n+90}的首项为-14-1+90=75
故Sn-n+90=75×(
5 |
6 |
5 |
6 |
由于Sn+1-Sn=1-
75 |
6 |
5 |
6 |