等比数列{an}的前n项和sn=a*2^n+a-2,则an=?

Sn=a*2^n+a-2,
Sn-1=a*2^n-1+a-2,
Sn-Sn-1=an=a*2^n-a*2^n-1=a*2^(n-1)
S1=a1=3a-2
S2=a1+a2=5a-2 a2=2a
S3=a1+a2+a3=9a-2 a3=4a
(2a)^2=4a*(3a-2) 解得 a=1
an=2^(n-1)