解：a1=2a1-3,a1=3 S(n-1)=2a(n-1)-3(n-1) (1) Sn=2an-3n (2) (2)-(1):an=2an-2a(n-1)-3, an=2a(n-1)+3 (an+3)=2(a(n-1)+3)所以｛an+3｝是一个等比数列,q=2. 令｛an+3｝=bn,b1=a1+3=6, bn=3×2^nan=bn-3=3×2^n－3
