已知数列{an}的前n项和为Sn,且满足Sn=n2+n(n∈N*).(I)求数列{an}的通项公式;(Ⅱ)若bn=2(n+1)an,求数列{bn}的前n项和Tn;(Ⅲ)若∃n∈N*,使Tn<C成立,求

题目简介

已知数列{an}的前n项和为Sn,且满足Sn=n2+n(n∈N*).(I)求数列{an}的通项公式;(Ⅱ)若bn=2(n+1)an,求数列{bn}的前n项和Tn;(Ⅲ)若∃n∈N*,使Tn<C成立,求

题目详情

已知数列{an}的前n项和为Sn,且满足Sn=n2+n(n∈N*)
(I)求数列{an}的通项公式;
(Ⅱ)若bn=
2
(n+1)an
,求数列{bn}的前n项和Tn
(Ⅲ)若∃n∈N*,使Tn<C成立,求实数C的取值范围.
题型:解答题难度:中档来源:不详

答案

(I)当n=1时,a1=S1=1+n=2.
当n≥2时,an=Sn-Sn-1=n2+n-[(n-1)2+n-1]=2n.当n=1时也成立.
an=2n(n∈N*)
(II)∵bn=class="stub"2
(n+1)an
=class="stub"2
2n(n+1)
=class="stub"1
n
-class="stub"1
n+1

∴数列{bn}的前n项和Tn=(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+
…+(class="stub"1
n
-class="stub"1
n+1
)

=1-class="stub"1
n+1

=class="stub"n
n+1

(III)若∃n∈N*,使Tn<C成立⇔(Tn)min<C,
∵n≥1,Tn=1-class="stub"1
n+1
≥1-class="stub"1
1+1
=class="stub"1
2
,即(Tn)min=class="stub"1
2

∴实数C的取值范围是(class="stub"1
2
,+∞)

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