在数列{an}中,前n项和为Sn,且Sn=n(n+1)2.(Ⅰ)求数列{an}的通项公式;(Ⅱ)设bn=an2n,数列{bn}前n项和为Tn,求Tn的取值范围.-数学

题目简介

在数列{an}中,前n项和为Sn,且Sn=n(n+1)2.(Ⅰ)求数列{an}的通项公式;(Ⅱ)设bn=an2n,数列{bn}前n项和为Tn,求Tn的取值范围.-数学

题目详情

在数列{an}中,前n项和为Sn,且Sn=
n(n+1)
2

(Ⅰ)求数列{an}的通项公式;
(Ⅱ)设bn=
an
2n
,数列{bn}前n项和为Tn,求Tn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;
当n≥2时,an=Sn-Sn-1=
n(n+1)
2
-
(n-1)n
2
=n
,经验证,a1=1满足上式.
故数列{an}的通项公式an=n.
(Ⅱ)可知Tn=class="stub"1
2
+class="stub"2
22
+class="stub"3
23
+…+class="stub"n
2n

class="stub"1
2
Tn=class="stub"1
22
+class="stub"2
23
+class="stub"3
24
+…+class="stub"n
2n+1

两式相减,得Tn-class="stub"1
2
Tn=class="stub"1
2
+class="stub"1
22
+class="stub"1
23
+…+class="stub"1
2n
-class="stub"n
2n+1
=1-class="stub"1
2n
-class="stub"n
2n+1

Tn=2-class="stub"n+2
2n

由于Tn+1-Tn=class="stub"n+1
2n+1
>0
,则Tn单调递增,故TnT1=class="stub"1
2

Tn=2-class="stub"n+2
2n
<2

故Tn的取值范围是[class="stub"1
2
,2)

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