解:因为 所以a1+3d>3,3a2≤9⇒d>2/ 3 ,a1+d≤3⇒a1≤3-d<3-2 /3 ="7" /3. ∵等差数列{an}的首项a1及公差d都是整数 ∴a1="2" 则1/ 3 <d≤1⇒d=1. ∴an=2+1×(n-1)=n+1. ∴bn=2nan=2n(n+1) 令Sn=b1+b2+…+bn =2•21+3•22+…+n•2n-1+(n+1)•2n① ∴2Sn=2•22+3•23+…+n•2n+(n+1)2n+1② ①-②得,-Sn=2•21+22+…+2n-(n+1)•2n+1=-n•2n+1-4 ∴Sn= 故答案为: |