已知数列{an}的前n项和Sn,满足:Sn=2an-2n(n∈N*).(1)求数列{an}的通项an;(2)若数列{bn}的满足bn=log2(an+2),Tn为数列{bnan+2}的前n项和,求证:

题目简介

已知数列{an}的前n项和Sn,满足:Sn=2an-2n(n∈N*).(1)求数列{an}的通项an;(2)若数列{bn}的满足bn=log2(an+2),Tn为数列{bnan+2}的前n项和,求证:

题目详情

已知数列{an}的前n项和Sn,满足:Sn=2an-2n(n∈N*)
(1)求数列{an}的通项an
(2)若数列{bn}的满足bn=log2(an+2),Tn为数列{
bn
an+2
}
的前n项和,求证:Tn
1
2
题型:解答题难度:中档来源:不详

答案

(1)当n∈N*时,Sn=2an-2n①,则当n≥2时,Sn-1=2an-1-2(n-1)②,
①-②,得an=2an-2an-1-2,即an=2an-1+2,
∴an+2=2(an-1+2),∴
an+2
an-1+2
=2

当n=1时,S1=2a1-2,则a1=2.
∴{an+2}是以a1+2=4为首项,2为公比的等比数列,
an+2=4•2n-1,∴an=2n+1-2
(2)证明:bn=log2(an+2)=log22n+1=n+1,∴
bn
an+2
=class="stub"n+1
2n+1

Tn=class="stub"2
22
+class="stub"3
23
+…+class="stub"n+1
2n+1
③,
class="stub"1
2
Tn=class="stub"2
23
+class="stub"3
24
+…+class="stub"n
2n+1
+class="stub"n+1
2n+2
…④,
③-④,得class="stub"1
2
Tn=class="stub"2
22
+class="stub"1
23
+class="stub"1
24
+…
+class="stub"1
2n+1
-class="stub"n+1
2n+2
=class="stub"1
2
+
class="stub"1
23
(1-class="stub"1
2n-1
)
1-class="stub"1
2
-class="stub"n+1
2n+2
=class="stub"3
4
-class="stub"n+3
2n+2

∴Tn=class="stub"3
2
-class="stub"n+3
2n+1

当n≥2时,Tn-Tn-1=-class="stub"n+3
2n+1
+class="stub"n+2
2n
=class="stub"n+1
2n+1
>0

∴{Tn}为递增数列,∴TnT1=class="stub"1
2

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