得a1=2a1+3-12,解得a1=9,
当n≥2时,an=Sn-Sn-1=(2an+3n-12)-[2an-1+3(n-1)-12]=2an-2an-1+3
∴an-3=2an-1-6=2(an-1-3),
∵a1-3=9-3=6,
∴{an-3}是首项为6,公比为2的等比数列.
∴an-3=6•2n-1,
∴an=6•2n-1+3=3(2n+1).
(Ⅱ)∵bn=nan=3n(2n+1)
∴Tn=3(1×2+2×22+3×23+…+n×2n)+
3n(n+1) |
2 |
设A=1×2+2×22+3×23+…+n×2n,
2A=1×22+2×23+3×24+…+n×2n+2,
两式相减,得:A=n×2n+1-(2+22+23+…+2n)
=n×2n+1-
2(1−2n) |
1−2 |
=(n-1)•2n+1+2,代入①式得
Tn=3[(n-1)•2n+1+2)+
3n(n+1) |
2 |
=3(n-1)•2n+1+
3n(n+1) |
2 |