已知{an}是首项为a1=1的等差数列且满足an+1>an(n∈N*),等比数列{bn}的前三项分别为b1=a1+1,b2=a2+1,b3=a3+3.(Ⅰ)求数列{an}和{bn}的通项公式;(Ⅱ)若

题目简介

已知{an}是首项为a1=1的等差数列且满足an+1>an(n∈N*),等比数列{bn}的前三项分别为b1=a1+1,b2=a2+1,b3=a3+3.(Ⅰ)求数列{an}和{bn}的通项公式;(Ⅱ)若

题目详情

已知{an}是首项为a1=1的等差数列且满足an+1>an(n∈N*),等比数列{bn}的前三项分别为b1=a1+1,b2=a2+1,b3=a3+3.
(Ⅰ)求数列{an}和{bn}的通项公式;
(Ⅱ)若数列{cn}满足(an+3)cnlog2bn=
1
2
,求数列{cn}的前n项和Sn
题型:解答题难度:中档来源:不详

答案

(Ⅰ)设等差数列{an}的公差为d,
首项a1=1,b1=2,b2=2+d,b3=4+2d,
∵{bn}为等比数列,∴
b22
=b1b3

即(2+d)2=2(4+2d),解得d=±2,
又∵an+1>an,即数列{an}为单调递增数列,
∴d=2,a2=3,a3=5,∴an=a1+(n-1)d=2n-1,
则b1=2,b2=4,q=2,
bn=b1qn-1=2n
∴an=2n-1,bn=2n
(Ⅱ)由题意得,(an+3)cnlog2bn=class="stub"1
2
,再由(1)结果代入,
变形得cn=class="stub"1
2(an+3)log2bn
=class="stub"1
2n(2n+2)
=class="stub"1
2
(class="stub"1
2n
-class="stub"1
2n+2
)

∴Sn=class="stub"1
2
(class="stub"1
2
-class="stub"1
4
)+class="stub"1
2
(class="stub"1
4
-class="stub"1
6
)+class="stub"1
2
(class="stub"1
6
-class="stub"1
8
)
+…+class="stub"1
2
(class="stub"1
2n
-class="stub"1
2n+2
)

=class="stub"1
2
(class="stub"1
2
-class="stub"1
2n+2
)
=class="stub"n
4(n+1)

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