数列{an}中,a1=1,Sn表示前n项的和,且Sn,Sn+1,2S1成等差数列

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数列{an}中,a1=1,Sn表示前n项的和,且Sn,Sn+1,2S1成等差数列
已知数列｛an｝中,a1=1,且Sn,Sn+1,2S1成等差数列,求此数列的前n项和公式Sn.

Sn,S(n+1),2S1成等差数列
Sn + 2S1= 2S(n+1)
Sn + 2 = 2S(n+1)
(S(n+1)-2) = (1/2)(Sn -2)
= (1/2)^n .(S1-2)
= -(1/2)^n
Sn -2 = -(1/2)^(n-1)
Sn = 2-(1/2)^(n-1)
再问：答案是Sn = 2- 1/（2^(n-1))，第4步没看懂
再答： Sn + 2S1= 2S(n+1) Sn + 2 = 2S(n+1) (S(n+1)-2) = (1/2)(Sn -2) {Sn -2} 是等比数列， q=1/2 (S(n+1)-2) = (1/2)(Sn -2) = (1/2)^n .(S1-2) = -(1/2)^n (S1=a1=1) Sn -2 = -(1/2)^(n-1) (代入 n'= n+1 ) Sn = 2-(1/2)^(n-1) (两边-2)
再问：本人愚笨= (1/2)(Sn -2) = (1/2)^n .(S1-2)这步不懂
再答： (S(n+1)-2) = (1/2)(Sn -2) (S(n+1)-2) /(Sn -2) = 1/2 令 bn = Sn -2 bn是等比数列， q=1/2 (S(n+1)-2) /(Sn -2) = 1/2 b(n+1) /bn =1/2 b(n+1)/b1 =(1/2)^n b(n+1) = (1/2)^n . b1 S(n+1) -2 = (1/2)^n . (S1-2)

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