已知Sn为数列{an}的前n项和,Sn=12n2+112n;数列满足:b3=11,bn+2=2bn+1-bn,其前9项和为153(1){bn}的通项公式;(2)设Tn为数列{cn}的前n项和,cn=6

题目简介

已知Sn为数列{an}的前n项和,Sn=12n2+112n;数列满足:b3=11,bn+2=2bn+1-bn,其前9项和为153(1){bn}的通项公式;(2)设Tn为数列{cn}的前n项和,cn=6

题目详情

已知Sn为数列{an}的前n项和,Sn=
1
2
n2+
11
2
n
;数列满足:b3=11,bn+2=2bn+1-bn,其前9项和为153
(1){bn}的通项公式;
(2)设Tn为数列{cn}的前n项和,cn=
6
(2an-11)(2bn-1)
,求使不等式T n
k
57
对∀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:不详

答案

(1)∵Sn=class="stub"1
2
n2+class="stub"11
2
n
,∴当n=1时,a1=S1=6;
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
经验证,当n=1时,上式也适合,
∴an=n+5;
∵bn+2=2bn+1-bn,∴bn+1=
bn+bn+2
2

∴{bn}是等差数列,设其公差为d.
b1+2d=11
9b1+36d=153
解得
b1=5
d=3

∴bn=5+3(n-1)=3n+2.
(2)∵cn=class="stub"6
(2an-11)(2bn-1)
=class="stub"6
[2(n+5)-11][2(3n+2)-1]

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

∴Tn=(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
)=1-class="stub"1
2n+1

∵n∈N+,∴Tn是单调递增数列,
∴当n=1时,(Tn)min=T1=1-class="stub"1
3
=class="stub"2
3

Tn>class="stub"k
57
对∀n∈N+都成立,等价于(Tn)min>class="stub"k
57
成立,
class="stub"2
3
>class="stub"k
57
,解得k<38
∴所求最大正整数k的值为37.

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