正项数列{an}中,前n项和为Sn,且a1=2,且an=22Sn-1+2(n≥2).(1)求数列{an}的通项公式;(2)设bn=an+82n+1,Tn=b1+b2+…+bn,证明52≤Tn<7.-数

题目简介

正项数列{an}中,前n项和为Sn,且a1=2,且an=22Sn-1+2(n≥2).(1)求数列{an}的通项公式;(2)设bn=an+82n+1,Tn=b1+b2+…+bn,证明52≤Tn<7.-数

题目详情

正项数列{an}中,前n项和为Sn,且a1=2,且an=2
2Sn-1
+2(n≥2)

(1)求数列{an}的通项公式;
(2)设bn=
an+8
2n+1
,Tn=b1+b2+…+bn,证明
5
2
Tn<7
题型:解答题难度:中档来源:不详

答案

(1)由an=2
2Sn-1
+2(n≥2)
,得Sn-Sn-1=2
2Sn-1
+2(n≥2)

Sn=Sn-1+2
2
Sn-1
+2=(
Sn-1
+
2
)2

Sn
=
Sn-1
+
2

{
Sn
}
是首项为
2
公差为
2
的等差数列,∴
Sn
=
2
n
,∴Sn=2n2
an=2
4(n-1)2
+2=4n-2(n≥2)
,对n=1也成立,
∴an=4n-2;
(2)证明:bn=class="stub"2n+3
2n

Tn=class="stub"5
21
+class="stub"7
22
+class="stub"9
23
+…+class="stub"2n+3
2n

class="stub"1
2
Tn=class="stub"5
22
+class="stub"7
23
+class="stub"9
24
+…+class="stub"2n+1
2n
+class="stub"2n+3
2n+1

两式相减,得class="stub"1
2
Tn=class="stub"5
2
+class="stub"2
22
+class="stub"2
23
+…+class="stub"2
2n
-class="stub"2
2n+1
=class="stub"7
2
-class="stub"2n+7
2n+1

所以T n=7-class="stub"2n+7
2n

n∈N∴class="stub"2n+7
2n
>0∴Tn<7

下面证明Tn≥class="stub"5
2

Tn+1-Tn=class="stub"2n+7
2n
-class="stub"2n+9
2n+1
=class="stub"2n+5
2n+1
>0
,∴Tn+1>Tn,∴{Tn}单调递增,
TnT1=class="stub"5
2

class="stub"5
2
Tn<7

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