已知各项均为正数的数列{an}前n项和为Sn,首项为a1,且12,an,Sn成等差数列.(1)求数列{an}的通项公式;(2)若an2=(12)bn,设cn=bnan,求数列{cn}的前n项和Tn.-

题目简介

已知各项均为正数的数列{an}前n项和为Sn,首项为a1,且12,an,Sn成等差数列.(1)求数列{an}的通项公式;(2)若an2=(12)bn,设cn=bnan,求数列{cn}的前n项和Tn.-

题目详情

已知各项均为正数的数列{an}前n项和为Sn,首项为a1,且
1
2
,an,Sn成等差数列.
(1)求数列{an}的通项公式;
(2)若an2=(
1
2
bn,设cn=
bn
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

(本小题满分12分)
(Ⅰ)由题意知2an=Sn+class="stub"1
2
an>0
,…(1分)
当n=1时,2a1=a1+class="stub"1
2
,解得a1=class="stub"1
2

当n≥2时,Sn=2an-class="stub"1
2
Sn-1=2an-1-class="stub"1
2

两式相减得an=Sn-Sn-1=2an-2an-1…(3分)
整理得:
an
an-1
=2
…(4分)
∴数列{an}是以class="stub"1
2
为首项,2为公比的等比数列.
an=a12n-1=class="stub"1
2
×2n-1=2n-2
.…(5分)
(Ⅱ)
a2n
=2-bn=22n-4

∴bn=4-2n,…(6分)
Cn=
bn
an
=class="stub"4-2n
2n-2
=class="stub"16-8n
2n
Tn=class="stub"8
2
+class="stub"0
22
+class="stub"-8
23
+…class="stub"24-8n
2n-1
+class="stub"16-8n
2n
…①
class="stub"1
2
Tn=class="stub"8
22
+class="stub"0
23
+…+class="stub"24-8n
2n
+class="stub"16-8n
2n+1
…②
①-②得class="stub"1
2
Tn=4-8(class="stub"1
22
+class="stub"1
23
+…+class="stub"1
2n
)-class="stub"16-8n
2n+1
…(9分)
=4-8•
class="stub"1
22
(1-class="stub"1
2n-1
)
1-class="stub"1
2
-class="stub"16-8n
2n+1
=4-4(1-class="stub"1
2n-1
)-class="stub"16-8n
2n+1
=class="stub"4n
2n
.…(11分)
Tn=class="stub"8n
2n
.…(12分)

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