问题 解答题
已知数列{an}满足:a1=2t-3(t∈R且t≠±1),an+1=
(2tn+1-3)an+2(t-1)tn-1
an+2tn-1
(n∈N*).
(1)当t=2时,求证:{
2n-1
an+1
}
是等差数列;
(2)若t>0,试比较an+1与an的大小;
(3)在(2)的条件下,已知函数f(x)=
x
x2+4
(x>0),是否存在正整数t,使得对一切n∈N*不等式f(an+1)<f(an)恒成立?若存在,求出t的最小值;若不存在,请说明理由.
答案

(1)证明:当t=2时,an+1=

(2n+2-3)an+2n+1-1
an+2n+1-1

an+1+1=

(2n+2-2)an+2n+2-2
an+2n+1-1

2n+1-1
an+1+1
=
an+2n+1-1
2(an+1)

2n+1-1
an+1+1
-
2n-1
an+1
=
1
2

{

2n-1
an+1
}是以
1
2
为公差的等差数列;

(2)∵an+1=

(2tn+1-3)an+2(t-1)tn-1
an+2tn-1
=
2(tn+1-1)(an+1)
an+2tn-1

an+1+1
tn+1-1
=
2(an+1)
an+2tn-1
=
2•
an+1
tn-1
an+1
tn-1
+2

an+1
tn-1
=bn,则bn+1=
2bn
bn+2
,b1=
a1+1
t-1
=2

1
bn-1
=
1
bn
+
1
2
1
b1
=
1
2

1
bn
=
n
2

an+1
tn-1
=
2
n

∴an=

2(tn-1)
n

∴an+1-an=

2(tn+1-1)
n+1
-
2(tn-1)
n
=
2(t-1)
n(n+1)
[n(1+t+…+tn)-(n+1)(1+t+…+tn-1)]

=

2(t-1)
n(n+1)
[(tn-1)+…+(tn-tn-1)]=
2(t-1)2
n(n+1)
[(1+t+…+tn-1)+t(1+t+…+tn-2)+…+tn-1]

显然t>0(t≠1)时,an+1-an>0,∴an+1>an

(3)∵f(an+1)-f(an)=

an+1
an+12+4
-
an
an2+4
=
(an+1-an)(an+1an-4)
(an+12+4)(an2+4
<0,an+1>an

∴an+1an-4>0,{an}为递增数列

∴只需a1a2-4>0

∴(2t-3)(t2-2)-4>0

令f(t)=(2t-3)(t2-2)-4,则f′(t)=6t2-6t-8

∴t>2时,f′(t)>0,函数为增函数

∵f(2)=-2<0,f(3)=17>0

∴满足题意的最小正整数t存在,最小值为3.

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