数列{an}前n项和为Sn,且2Sn+1=3an,求an及Sn

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数列{an}前n项和为Sn,且2Sn+1=3an,求an及Sn
数列{an}前n项和为Sn,且2Sn+1=3an,求an及Sn

数列{an}前n项和为Sn,且2Sn+1=3an,求an及Sn
当n=1时、有2s1+1=3a1,即有a1=1,因为2Sn+1=3an,所以2Sn+1+1=3an+1.后式减去前式,得2an+1=3an+1-3an.即有
an+1=3an,为等比数列,且公比为3,所以an=3^(n-1).
sn=3^n-1/2.

3A(n+1)=2S(n+1)+1 (1)
3An=2Sn+1 (2)
1式-2式得
3A(n+1)-3An=2A(n+1)
所以A(n+1)=3An
因为A1=1
所以An=3^(n-1) ,q=3.
Sn=a1*(q^n-1)/(q-1)=(3^n-1)/2

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