(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. |