题目内容
[已知数列{an}满足:a1=-
,a2=1,数列{
}为等差数列;数列{bn}中,Sn为其前n项和,且b1=
,4n•Sn+3n+1=3•4n.
(1)求证:数列{bn}是等比数列;
(2)记An=anan+1,求数列{An}的前n项和S;
(3)设数列{cn}满足cn=
,Tn为数列{cn}的前n项和,求xn=Tn+1-2Tn+Tn-1的最大值.
1 |
2 |
1 |
an |
3 |
4 |
(1)求证:数列{bn}是等比数列;
(2)记An=anan+1,求数列{An}的前n项和S;
(3)设数列{cn}满足cn=
bn |
an |
分析:(1)根据给出的数列{bn}的前n项和所满足的等式,求出Sn,然后由bn=
求出通项,继而可说明数列{bn}是等比数列;
(2)由数列{
}为等差数列求出数列{an}的通项公式,然后运用裂项法求数列{An}的前n项和S;
(3)把an,bn的通项公式代入求cn,把xn=Tn+1-2Tn+Tn-1变形后换上cn,得到关于n的函数式,写出Xn+1,与Xn作差后分析差式的单调性,从而得到Xn的最大值.
|
(2)由数列{
1 |
an |
(3)把an,bn的通项公式代入求cn,把xn=Tn+1-2Tn+Tn-1变形后换上cn,得到关于n的函数式,写出Xn+1,与Xn作差后分析差式的单调性,从而得到Xn的最大值.
解答:解:(1)由4n•Sn+3n+1=3•4n得,Sn=3-3•(
)n,当n≥2时,bn=Sn-Sn-1=(
)n,又b1=
,故bn=(
)n,故数列{bn}是等比数列;
(2)∵a1=-
,a2=1,∴
=-2,
=1,∴d=
-
=1-(-2)=3,∴
=-2+(n-1)•3=3n-5,则an=
,
∴An=
=
(
-
),
∴S=
[(-
-1)+(1-
)+(
-
)+…+(
-
)]=
(-
-
)=
;
(3)∵cn=(3n-5)•(
)n
∴xn=Tn+1-2Tn+Tn-1=(Tn+1-Tn)-(Tn-Tn-1)=cn+1-cn=(
)n(
),
xn+1-xn=(
)n+1(
)-(
)n(
)=(
)n(
),
故当n≤7时,{xn}是递减的,当n≥8时,{xn}是递增的,但n≥8时,xn<0
故xn的最大值为x1=(
)•(
)=
.
3 |
4 |
3 |
4 |
3 |
4 |
3 |
4 |
(2)∵a1=-
1 |
2 |
1 |
a1 |
1 |
a2 |
1 |
a2 |
1 |
a1 |
1 |
an |
1 |
3n-5 |
∴An=
1 |
(3n-5)(3n-2) |
1 |
3 |
1 |
3n-5 |
1 |
3n-2 |
∴S=
1 |
3 |
1 |
2 |
1 |
4 |
1 |
4 |
1 |
7 |
1 |
3n-5 |
1 |
3n-2 |
1 |
3 |
1 |
2 |
1 |
3n-2 |
-n |
6n-4 |
(3)∵cn=(3n-5)•(
3 |
4 |
∴xn=Tn+1-2Tn+Tn-1=(Tn+1-Tn)-(Tn-Tn-1)=cn+1-cn=(
3 |
4 |
14-3n |
4 |
xn+1-xn=(
3 |
4 |
11-3n |
4 |
3 |
4 |
14-3n |
4 |
3 |
4 |
3n-23 |
16 |
故当n≤7时,{xn}是递减的,当n≥8时,{xn}是递增的,但n≥8时,xn<0
故xn的最大值为x1=(
3 |
4 |
11 |
4 |
33 |
16 |
点评:本题是等差数列和等比数列的综合题,考查了裂项法对数列求和,(3)的解答运用函数思想,借助于函数的单调性分析出了函数取最大值时的n的值,该题是中档以上难度题型.
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