题目内容

设数列{an}的前n项积为Tn,已知对?n,m∈N+,当n>m时,总有
Tn
Tm
=Tn-mq(n-m)m
(q>0是常数).
(1)求证:数列{an}是等比数列;
(2)设正整数k,m,n(k<m<n)成等差数列,试比较Tn•Tk和(Tm2的大小,并说明理由;
(3)探究:命题p:“对?n,m∈N+,当n>m时,总有
Tn
Tm
=Tn-mq(n-m)m
(q>0是常数)”是命题t:“数列{an}是公比为q(q>0)的等比数列”的充要条件吗?若是,请给出证明;若不是,请说明理由.
分析:(1)设m=1,则有
Tn
T1
=Tn-1qn-1
,从而可得an=a1qn-1,即可证得数列{an}是等比数列;
(2)当q=1时,Tn•Tk=a1n+k=a12m=Tm2;当q≠1时,an=a1qn-1Tn=a1nq
n(n-1)
2
,从而可得Tn•Tk=a1nq
n(n-1)
2
a1kq
k(k-1)
2
=a1n+kq
n(n-1)+k(k-1)
2
,根据Tm2=a12mqm(m-1),n+k=2m,k<m<n,利用基本不等式,即可得到结论;
(3)证明:由(1)知,充分性成立;
必要性:利用q≠1时,an=a1qn-1Tn=a1nq
n(n-1)
2
,可证得
Tn
Tm
=Tn-mq(n-m)m
,同理可证,当q=1时,也成立,故得证.
解答:(1)证明:设m=1,则有
Tn
T1
=Tn-1qn-1
,∴
Tn
Tn-1
=a1qn-1

an=a1qn-1
∴n≥2时,
an
an-1
=q

∴数列{an}是等比数列;
(2)解:当q=1时,an=a1,∴Tn=a1n,∴Tn•Tk=a1n+k=a12m=Tm2
当q≠1时,an=a1qn-1Tn=a1nq
n(n-1)
2

∴Tn•Tk=a1nq
n(n-1)
2
a1kq
k(k-1)
2
=a1n+kq
n(n-1)+k(k-1)
2

Tm2=a12mqm(m-1),n+k=2m,k<m<n
a12m=a1n+k
n(n-1)+k(k-1)
2
=
n2+k2
2
-m
(
n+k
2
)
2
-m=m2-m

∴q>1时,Tn•TkTm2;q<1时,Tn•TkTm2
(3)证明:由(1)知,充分性成立;
必要性:若数列{an}是公比为q(q>0)的等比数列,则an=a1qn-1
∴q≠1时,Tn=a1nq
n(n-1)
2

Tn
Tm
=
a1nq
n(n-1)
2
a1mq
m(m-1)
2
=a1n-mq
(n-m)(n+m+1)
2

Tn-mq(n-m)m=a1n-mq
(n-m)(n-m-1)
2
•q(n-m)m=a1n-mq
(n-m)(n+m+1)
2

Tn
Tm
=Tn-mq(n-m)m

∴对?n,m∈N+,当n>m时,总有
Tn
Tm
=Tn-mq(n-m)m
(q>0是常数)
同理可证,当q=1时,也成立
∴命题p:“对?n,m∈N+,当n>m时,总有
Tn
Tm
=Tn-mq(n-m)m
(q>0是常数)”是命题t:“数列{an}是公比为q(q>0)的等比数列”的充要条件.
点评:本题考查等比数列的定义,考查新定义,考查充要性的证明,综合性强,难度大.
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