数列{an}的前n项和记作Sn,满足 Sn=2an+3n-12(n∈N*) (Ⅰ)证明数列{an-3}为等比数列,并求出数列{an}的通项公式; (Ⅱ)记bn=nan,数列{bn}的前n项和为Tn,求Tn.

问题描述:

数列{an}的前n项和记作Sn,满足 Sn=2an+3n-12(n∈N*
(Ⅰ)证明数列{an-3}为等比数列,并求出数列{an}的通项公式;
(Ⅱ)记bn=nan,数列{bn}的前n项和为Tn,求Tn

(Ⅰ)证明:把n=1代入Sn=2an+3n-12,
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
,①
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
+6.