(1)由a1=1,an+1=2sn,分别令n=1,2,3求出a2,a3,a4的值;
(2)由an+1=2sn及求得an,
(3)把(2)求得an代入中bn,应用错位相减法求和.
【解析】
(1)∵a1=1,
∴a2=2a1=2,a3=2S2=6,a4=2S3=18,
(2)∵an+1=2S1,∴aN=2Sn-1(n≥2),
∴an+1-an=2an,(n≥2)
又,∴数列{an}自第2项起是公比为3的等比数列,
∴,
(3)∵bn=nan,∴,
∴Tn=1+2×2×3+2×3×31+2×4×32++2×n×3n-2,①
3Tn=3+2×2×31+2×3×32+2×4×33++2×n×3n-1②(12分)
①-②得-2Tn=-2+2×2×3+2×31+2×32++2×3n-2-2×n×3n-1
=2+2(3+32+33++3n-2)-2n×3n-1=(1-2n)×3n-1-1
∴.(14分)