已知数列{an}的前n项的和Sn=n2+2n,数列{bn}是正项等比数列,且满足a1=2b1,b3(a3-a1)=b1.(Ⅰ)求数列{an}和{bn}的通项公式;(Ⅱ)记cn=an•bn,求数列{cn

题目简介

已知数列{an}的前n项的和Sn=n2+2n,数列{bn}是正项等比数列,且满足a1=2b1,b3(a3-a1)=b1.(Ⅰ)求数列{an}和{bn}的通项公式;(Ⅱ)记cn=an•bn,求数列{cn

题目详情

已知数列{an}的前n项的和Sn=n2+2n,数列{bn}是正项等比数列,且满足a1=2b1,b3(a3-a1)=b1
(Ⅰ)求数列{an}和{bn}的通项公式;
(Ⅱ)记cn=an•bn,求数列{cn}的前n项的和.
题型:解答题难度:中档来源:不详

答案

解(1)数列{an}前n项的和Sn=n2+2n∴an=Sn-Sn-1=2n+1(n∈N,n≥2)(2分)
又an=S1=3,
所以数列{an}的通项公式为an=2n+1(n∈N*)(3分)
因为数列{bn}是正项等比数列,b1=class="stub"1
2
a1=class="stub"3
2
a3-a1=4
,∴
b3
b1
=class="stub"1
a3-a1
=class="stub"1
4
,(4分)
公比为class="stub"1
2
,(5分)
数列{bn}的通项公式为bn=class="stub"3
2
•class="stub"1
2n-1
=3•(class="stub"1
2
)n(n∈N*)
(6分)
(2)所以cn=3(2n+1)(class="stub"1
2
)n
,设数列{cn}的前n项的和为TnTn=3[3•class="stub"1
2
+5•(class="stub"1
2
)2+
+(2n+1)•(class="stub"1
2
)n]

class="stub"1
2
Tn=3[3•(class="stub"1
2
)2+5•(class="stub"1
2
)3
+…+(2n-1)(class="stub"1
2
)
n
+(2n+1)(class="stub"1
2
)
n+1
]
(1-class="stub"1
2
)Tn=3{3•class="stub"1
2
+2[(class="stub"1
2
)2+(class="stub"1
2
)3+
…+(class="stub"1
2
)n]-(2n+1)•(class="stub"1
2
)n+1}

class="stub"1
2
Tn=3{3•class="stub"1
2
+2[
(class="stub"1
2
)
2
(1-(class="stub"1
2
)
n-1
)
1-class="stub"1
2
]-(2n+1)•(class="stub"1
2
)n+1}

Tn=15-(6n+15)•(class="stub"1
2
)n
(12分)

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