已知等差数列{an}为递增数列,且a2,a5是方程x2-12x+27=0的两根,数列{bn}的前n项和Tn=1-12bn.(1)求数列{an}和{bn}的通项公式.(2)若Cn=3nbnanan+1,

题目简介

已知等差数列{an}为递增数列,且a2,a5是方程x2-12x+27=0的两根,数列{bn}的前n项和Tn=1-12bn.(1)求数列{an}和{bn}的通项公式.(2)若Cn=3nbnanan+1,

题目详情

已知等差数列{an}为递增数列,且a2,a5是方程x2-12x+27=0的两根,数列{bn}的前n项和Tn=1-
1
2
bn
(1)求数列{an}和{bn}的通项公式.
(2)若Cn=
3nbn
anan+1
,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

(1)①∵等差数列{an}为递增数列,且a2,a5是方程x2-12x+27=0的两根,
∴a2+a5=12,a2a5=27,
∵d>0,∴a2=3,a5=9,
∴d=
a5-a2
3
=2,a1=1,
∴an=2n-1(n∈N*)
②∵Tn=1-class="stub"1
2
bn,
∴令n=1,得b1=class="stub"2
3

当n≥2时,Tn=1-class="stub"1
2
bn,Tn-1=1-class="stub"1
2
bn-1,两式相减得,bn=class="stub"1
2
bn-1-class="stub"1
2
bn,
bn
bn-1
=class="stub"1
3
(n≥2),
数列{bn}是以class="stub"2
3
为首项,class="stub"1
3
为公比的等比数列.
∴bn=class="stub"2
3
(class="stub"1
3
)n-1
=2•class="stub"1
3n
(n∈N*).
(2)∵bn=2•class="stub"1
3n
,Cn=
3nbn
anan+1

∴Cn=
3n×2×class="stub"1
3n
(2n-1)(2n+1)
=class="stub"1
2n-1
-class="stub"1
2n+1

∴Sn=(1-class="stub"1
3
)+(class="stub"1
3
-class="stub"1
5
)+
…+(class="stub"1
2n-1
-class="stub"1
2n+1
)
=1-class="stub"1
2n+1
=class="stub"2n
2n+1

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