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)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
=
+
;方法
令bn=
,则数列{bn}是首项b1=1、公差为
的等差数列,bn=
;
∴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.
故Tn=3×[2n+1(n-1)+2]-4
=2n+1(3n-7)+14•
当n≥2时,an=Sn-Sn-1,an=2an-1+3×2n-1,于是
an |
2n |
an-1 |
2n-1 |
3 |
2 |
令bn=
an |
2n |
3 |
2 |
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.
故Tn=3×[2n+1(n-1)+2]-4
2(1-2n) |
1-2 |
练习册系列答案
相关题目