设数列{an}的前n项和为sn,若对于任意的正整数n都有sn=2an-3n.(1)设bn=an+3,证明:数列｛bn｝是等比数列

(1)Sn=2an-3nn=1时,S1=a1,故有：a1=2a1-3,a1=3n>=2时,an=Sn-S(n-1)＝2an-3n-[2a(n-1)-3(n-1)]＝2an-2a(n-1)-3即：an＝2a(n-1)+3两边+3an+3＝2[a(n-1)+3]而bn=an+3,代入：bn=2b(n-1)所以数列{bn}是等比数列,q=2,首项...