已知数列{an}的前n项为和Sn,点(n,Snn)在直线y=12x+112上.数列{bn}满足bn+2-2bn+1+bn=0(n∈N*),且b3=11,前9项和为153.(Ⅰ)求数列{an}、{bn}

题目简介

已知数列{an}的前n项为和Sn,点(n,Snn)在直线y=12x+112上.数列{bn}满足bn+2-2bn+1+bn=0(n∈N*),且b3=11,前9项和为153.(Ⅰ)求数列{an}、{bn}

题目详情

已知数列{an}的前n项为和Sn,点(n,
Sn
n
)
在直线y=
1
2
x+
11
2
上.数列{bn}满足bn+2-2bn+1+bn=0(n∈N*),且b3=11,前9项和为153.
(Ⅰ)求数列{an}、{bn}的通项公式;
(Ⅱ)设cn=
3
(2an-11)(2bn-1)
,数列{cn}的前n和为Tn,求使不等式Tn
k
57
对一切n∈N*都成立的最大正整数k的值.
题型:解答题难度:中档来源:朝阳区一模

答案

(Ⅰ)由题意,得
Sn
n
=class="stub"1
2
n+class="stub"11
2
,即Sn=class="stub"1
2
n2+class="stub"11
2
n

故当n≥2时,an=Sn-Sn-1=(class="stub"1
2
n2+class="stub"11
2
n)-[class="stub"1
2
(n-1)2+class="stub"11
2
(n-1)]=n+5

注意到n=1时,a1=S1=6,而当n=1,n+5=6,
所以,an=n+5(n∈N*).
又bn+2-2bn+1+bn=0,即bn+2-bn+1=bn+1-bn(n∈N*),
所以{bn}为等差数列,于是
9(b3+b7)
2
=153

b3=11,故b7=23,d=class="stub"23-11
7-3
=3

因此,bn=b3+3(n-3)=3n+2,即bn=3n+2(n∈N*).
(Ⅱ)cn=class="stub"3
(2an-11)(2bn-1)
=class="stub"3
[2(n+5)-11][2(3n+2)-1]

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

所以,Tn=c1+c2+…+cn=class="stub"1
2
[(1-class="stub"1
3
)+(class="stub"1
3
-class="stub"1
5
)+(class="stub"1
5
-class="stub"1
7
)++(class="stub"1
2n-1
-class="stub"1
2n+1
)]

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

由于Tn+1-Tn=class="stub"n+1
2n+3
-class="stub"n
2n+1
=class="stub"1
(2n+3)(2n+1)
>0

因此Tn单调递增,故(Tn)min=class="stub"1
3

class="stub"1
3
>class="stub"k
57
,得k<19,所以Kmax=18

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