设数列{an}前n项和Sn,且Sn=2an-2,n∈N+.(Ⅰ)试求数列{an}的通项公式;(Ⅱ)设cn=nan,求数列{cn}的前n项和Tn.-数学

题目简介

设数列{an}前n项和Sn,且Sn=2an-2,n∈N+.(Ⅰ)试求数列{an}的通项公式;(Ⅱ)设cn=nan,求数列{cn}的前n项和Tn.-数学

题目详情

设数列{an}前n项和Sn,且Sn=2an-2,n∈N+
(Ⅰ)试求数列{an}的通项公式;
(Ⅱ)设cn=
n
an
,求数列{cn}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(Ⅰ)当n≥2时,an=Sn-Sn-1=(2an-2)-(2an-1-2)=2an-2an-1,
所以,an=2an-1,即
an
an-1
=2
,…(3分)
当n=1时,S1=2a1-2,a1=2,…(4分)
由等比数列的定义知,数列{an}是首项为2,公比为2的等比数列,
所以,数列{an}的通项公式为an=2×2n-1=2n,n∈N+.…(6分)
(II)由(I)知,cn=class="stub"n
an
=class="stub"n
2n
(8分)
Tn=class="stub"1
2
+class="stub"2
22
+…+class="stub"n-1
2n-1
+class="stub"n
2n

class="stub"1
2
Tn
=class="stub"1
22
+class="stub"2
23
+…+class="stub"n-1
2n
+class="stub"n
2n+1

两式相减可得,class="stub"1
2
Tn
=class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2n
-class="stub"n
2n+1
=
class="stub"1
2
(1-class="stub"1
2n
)
1-class="stub"1
2
-class="stub"n
2n+1
=1-class="stub"1
2n
-class="stub"n
2n+1

∴Tn=2-class="stub"n+2
2n
(12分)

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