已知各项均为正数的数列{an}的前n项和为Sn,且Sn、an、12成等差数列.(Ⅰ)求数列{an}的通项公式;(Ⅱ)若a2n=2-bn,设Cn=bnan,求数列{Cn}的前项和Tn.-数学

题目简介

已知各项均为正数的数列{an}的前n项和为Sn,且Sn、an、12成等差数列.(Ⅰ)求数列{an}的通项公式;(Ⅱ)若a2n=2-bn,设Cn=bnan,求数列{Cn}的前项和Tn.-数学

题目详情

已知各项均为正数的数列{an}的前n项和为Sn,且Sn、an
1
2
成等差数列.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)若
a2n
=2-bn
,设Cn=
bn
an
,求数列{Cn}的前项和Tn
题型:解答题难度:中档来源:湖北模拟

答案

(Ⅰ) 由题意知2an=Sn+class="stub"1
2
an>0

当n=1时,2a1=a1+class="stub"1
2
a1=class="stub"1
2

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

两式相减得an=2an-2an-1(n≥2),整理得:
an
an-1
=2
(n≥2)
∴数列{an}是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

(Ⅱ)
a2n
=2-bn=22n-4

∴bn=4-2n
Cn=
ba
aa
=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
class="stub"1
2
Tn=class="stub"8
22
+class="stub"0
23
+…+class="stub"24-8n
2n
+class="stub"16-8n
2n+1

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

=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

Tn=class="stub"8n
2n

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