(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 |
∴=
∴-=
∴{}是以为公差的等差数列;
(2)∵an+1=(2tn+1-3)an+2(t-1)tn-1 |
an+2tn-1 |
=
∴==
令=bn,则bn+1=,b1==2
∴=+,=
∴=
∴=
∴an=
∴an+1-an=-=[n(1+t+…+tn)-(n+1)(1+t+…+tn-1)]
=[(tn-1)+…+(tn-tn-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)(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.