(1)由题意可知:Sn-1=1- (n≥2), 又2n-1·an=Sn-Sn-1,∴2n-1·an=-. ∴an=-=-2-n(n≥2).∴a1=-. 又S1=1-=,∴a1≠S1,∴an= (2)由题意知bn= (n≥2),∴=n·2n(n≥2). ∵==2,∴=n·2n(n≥1). 设的前n项和为,则=1×2+2×22+3×23+…+n·2n, 2=1×22+2×23+3×24+…+(n-1)·2n+n·2n+1, ∴-2=1×2+22+23+…+2n-n·2n+1=2+22+…+2n-n·2n+1, ∴-=(1-n)·2n+1-2,∴=(n-1)·2n+1+2 |