题目内容

已知数列{an}的前n项和Sn=2an-3•2n+4,n=1,2,3,….
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)设Tn为数列{Sn-4}的前n项和,求Tn
(Ⅰ)∵a1=S1=2a1-2,∴a1=2.
当n≥2时,an=Sn-Sn-1,an=2an-1+3×2n-1,于是
an
2n
=
an-1
2n-1
+
3
2
;方法
bn=
an
2n
,则数列{bn}是首项b1=1、公差为
3
2
的等差数列,bn=
3n-1
2

∴an=2nbn=2n-1(3n-1).
(Ⅱ)∵Sn-4=2n(3n-4)=3×2n×n-2n+2
∴Tn=3(2×1+22×2++2n×n)-4(2+22++2n),
记Wn=2×1+22×2++2n×n①,则2Wn=22×1+23×2++2n+1×n②,
①-②有-Wn=2×1+22++2n-2n+1×n=2n+1(1-n)-2,
∴Wn=2n+1(n-1)+2.
Tn=3×[2n+1(n-1)+2]-4
2(1-2n)
1-2
=2n+1(3n-7)+14•
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