题目内容
设首项为a1的正项数列{an}的前n项和为Sn,q为非零常数,已知对任意正整数n,m,Sn+m=Sm+qmSn总成立.(Ⅰ)求证:数列{an}是等比数列;
(Ⅱ)若不等的正整数m,k,h成等差数列,试比较amm•ahh与ak2k的大小;
(Ⅲ)若不等的正整数m,k,h成等比数列,试比较
a |
m |
a |
h |
a |
k |
分析:(Ⅰ)令n=m=1,得a2=qa1,令m=1,得Sn+1=S1+qSn(1),从而Sn+2=S1+qSn+1两式相减即可得出an+2=qan+1,进而可判断出数列{an}是等比数列
(Ⅱ)根据m,k,h成等差数列,可知m+h=2k,进而可判定m2+h2>
(m+h)2=2k2,进而根据等比数列的通项公式分q大于、等于和小于1三种情况判断.
(Ⅲ)正整数m,k,h成等比数列,则m•h=k2,判断出
+
>2
=
,进而根据等差根据等比数列的通项公式分a1和q大于、等于和小于1三种情况判断.
(Ⅱ)根据m,k,h成等差数列,可知m+h=2k,进而可判定m2+h2>
1 |
2 |
(Ⅲ)正整数m,k,h成等比数列,则m•h=k2,判断出
1 |
m |
1 |
h |
|
2 |
k |
解答:(Ⅰ)证:因为对任意正整数n,m,Sn+m=Sm+qmSn总成立,
令n=m=1,得S2=S1+qS1,则a2=qa1
令m=1,得Sn+1=S1+qSn(1),从而Sn+2=S1+qSn+1(2),
(2)-(1)得an+2=qan+1,(n≥1)
综上得an+1=qan(n≥1),所以数列{an}是等比数列
(Ⅱ)正整数m,k,h成等差数列,
则m+h=2k,
所以m2+h2>
(m+h)2=2k2,
则
•
=
qm2-m
qh2-h=
qm2+h2-m-h
①当q=1时,amm•ahh=a12k=ak2k
②当q>1时,
•
=
qm2+h2-m-h>
q2k2-2k=(a1qk-1)2k=
③当0<q<1时,
•
=
qm2+h2-m-h<
q2k2-2k=(a1qk-1)2k=
(Ⅲ)正整数m,k,h成等比数列,则m•h=k2,则
+
>2
=
,
所以
•
=(a1qm-1)
(a1qh-1)
=
q2-
-
=q2(
)
+
,ak
=q2(
)
①当a1=q,即
=1时,
•
=
=q2=ak
②当a1>q,即
>1时,
•
=q2(
)
+
>q2(
)
=ak
③当a1<q,即
<1时,
•
=q2(
)
+
<q2(
)
=ak
令n=m=1,得S2=S1+qS1,则a2=qa1
令m=1,得Sn+1=S1+qSn(1),从而Sn+2=S1+qSn+1(2),
(2)-(1)得an+2=qan+1,(n≥1)
综上得an+1=qan(n≥1),所以数列{an}是等比数列
(Ⅱ)正整数m,k,h成等差数列,
则m+h=2k,
所以m2+h2>
1 |
2 |
则
a | m m |
a | h h |
a | m 1 |
a | h 1 |
a | 2k 1 |
①当q=1时,amm•ahh=a12k=ak2k
②当q>1时,
a | m m |
a | h h |
a | 2k 1 |
a | 2k 1 |
a | 2k k |
③当0<q<1时,
a | m m |
a | h h |
a | 2k 1 |
a | 2k 1 |
a | 2k k |
(Ⅲ)正整数m,k,h成等比数列,则m•h=k2,则
1 |
m |
1 |
h |
|
2 |
k |
所以
a |
m |
a |
h |
1 |
m |
1 |
h |
a |
1 |
1 |
m |
1 |
h |
a1 |
q |
1 |
m |
1 |
h |
2 |
k |
a1 |
q |
2 |
k |
①当a1=q,即
a1 |
q |
a |
m |
a |
h |
a |
k |
2 |
k |
②当a1>q,即
a1 |
q |
a |
m |
a |
h |
a1 |
q |
1 |
m |
1 |
h |
a1 |
q |
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2 |
k |
③当a1<q,即
a1 |
q |
a |
m |
a |
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a1 |
q |
1 |
m |
1 |
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a1 |
q |
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点评:本题主要考查了等比关系的确定和等比数列的性质.等比数列常与不等式一块考查,应引起重视.

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