已知数列{an}的前n项和为Sn,且有Sn=12n2+112n,数列{bn}满足bn+2-2bn+1+bn=0(n∈N*),且b3=11,前9项和为153;(1)求数列{an}的通项公式;(2)求数列

题目简介

已知数列{an}的前n项和为Sn,且有Sn=12n2+112n,数列{bn}满足bn+2-2bn+1+bn=0(n∈N*),且b3=11,前9项和为153;(1)求数列{an}的通项公式;(2)求数列

题目详情

已知数列{an}的前n项和为Sn,且有Sn=
1
2
n2+
11
2
n,数列{bn}满足bn+2-2bn+1+bn=0(n∈N*),且b3=11,前9项和为153;
(1)求数列{an}的通项公式;
(2)求数列{bn}的通项公式;
(3)设cn=
3
(2an-11)(2bn-1)
,数列{cn}的前n项和为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:不详

答案

(1)因为Sn=class="stub"1
2
n2+class="stub"11
2
n,故
当n≥2时,an=Sn-Sn-1=n+5;当n=11时,a1=S1=6;满足上式;
所以an=n+5,
(2)又因为bn+2-2bn+1+bn=0,所以数列{bn}为等差数列;
由S9=
9(b3+b7)
2
=153,b3=11,故b7=23;所以公差d=class="stub"23-11
7-3
=3;
所以:bn=b3+(n-3)d=3n+2;
(3)由(1)知:Cn=class="stub"3
(2an-11)(2bn-1)
=class="stub"1
(2n-1)(2n+1)

而Cn=class="stub"3
(2an-11)(2bn-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+c3+c4+…+cn=class="stub"1
2
[1-class="stub"1
3
+class="stub"1
3
-class="stub"1
5
+…+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

又因为Tn+1-Tn=class="stub"n+1
2n+3
-class="stub"n
2n+1
=class="stub"1
(2n+3)(2n+1)
>0;
所以{Tn}是单调递增,故(Tn)min=T1=class="stub"1
3

由题意可知class="stub"1
3
class="stub"k
57
;得k<19,所以k的最大正整数为18;

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