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
1
m
m
a
1
h
h
a
2
k
k
的大小.
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
1
2
(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
1
m
+
1
h
>2
1
mh
=
2
k
,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.
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
1
2
(m+h)2=2k2

a
m
m
a
h
h
=
a
m
1
qm2-m
a
h
1
qh2-h=
a
2k
1
qm2+h2-m-h

①当q=1时,amm•ahh=a12k=ak2k
②当q>1时,
a
m
m
a
h
h
=
a
2k
1
qm2+h2-m-h
a
2k
1
q2k2-2k=(a1qk-1)2k=
a
2k
k

③当0<q<1时,
a
m
m
a
h
h
=
a
2k
1
qm2+h2-m-h
a
2k
1
q2k2-2k=(a1qk-1)2k=
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
1
mh
=
2
k

所以
a
1
m
m
a
1
h
h
=(a1qm-1)
1
m
(a1qh-1)
1
h
=
a
1
m
+
1
h
1
q2-
1
m
-
1
h
=q2(
a1
q
)
1
m
+
1
h
ak
2
k
=q2(
a1
q
)
2
k

①当a1=q,即
a1
q
=1
时,
a
1
m
m
a
1
h
h
=
a
2
k
k
=q2=ak
2
k

②当a1>q,即
a1
q
>1
时,
a
1
m
m
a
1
h
h
=q2(
a1
q
)
1
m
+
1
h
q2(
a1
q
)
2
k
=ak
2
k

③当a1<q,即
a1
q
<1
时,
a
1
m
m
a
1
h
h
=q2(
a1
q
)
1
m
+
1
h
q2(
a1
q
)
2
k
=ak
2
k
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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