已知各项为正数的数列{an}的前n项和为{Sn},首项为a1,且2,an,Sn成等差数列,(Ⅰ)求数列{an}的通项公式;(Ⅱ)若bn=log2an,cn=bnan,求数列{cn}的前n项和Tn.-数

题目简介

已知各项为正数的数列{an}的前n项和为{Sn},首项为a1,且2,an,Sn成等差数列,(Ⅰ)求数列{an}的通项公式;(Ⅱ)若bn=log2an,cn=bnan,求数列{cn}的前n项和Tn.-数

题目详情

已知各项为正数的数列{an}的前n项和为{Sn},首项为a1,且2,an,Sn成等差数列,
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)若bn=log2ancn=
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

(I)由题意可得,2an=2+Sn①
∴2an-1=2+Sn-1(n≥2)②
①-②可得,an=2an-1(n≥2)
∵2a1=2+S1∴a1=2
由等比数列的通项公式可得,an=2n
(II)∵bn=log2an=n,Cn=
bn
an
=class="stub"n
2n

∴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"n-1
2n
+class="stub"n
2n+1

①-②可得,class="stub"1
2
Tn
=class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2n
-class="stub"n
2n+1
=
class="stub"1
2
[1-(class="stub"1
2
)
n
]
1-class="stub"1
2
-class="stub"n
2n+1

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

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