Question

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。过程及答案。谢谢

Answer 1

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)

追问

答案是Sn = 2- 1/（2^(n-1))，第4步没看懂

追答

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)

追问

本人愚笨= (1/2)(Sn -2)
= (1/2)^n .(S1-2)这步不懂

追答

(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)