已知sn是数列{an}的前n项和,a1=3/2,a2=2,且S（n+1）-3Sn+(2Sn-1)+1=0,其中n大于等于

S(n+1)-3Sn+2S(n-1)+1=0
S(n+1)-Sn -1 = 2(Sn-S(n-1) - 1)
[S(n+1)-Sn -1]/(Sn-S(n-1) - 1) =2
[S(n+1)-Sn -1]/(S2-S1 - 1) = 2^(n-1)
S(n+1)-Sn -1 = 2^(n-1)
S(n+1)-Sn = 1+ 2^(n-1)
Sn -S(n-1) = 1+ 2^(n-2)
Sn - S2 = [1+ 2^1] +[1+ 2^2]+...+[1+ 2^(n-2)]
= (n-2)+ 2(2^(n-2) -1)
= n-4 + 2^(n-1)
Sn = n- 1/2 + 2^(n-1)
an = Sn-S(n-1)
= 1+ 2^(n-2)
an = 3/2 ; n=1
= 2 ; n=2
= 1+ 2^(n-2) ; n>=3
an -1 = 2^(n-2)
{an-1} 是等比数列