一道数列的题，有追分！

1.
n>=2时
An=Sn-S(n-1)
=(1/2)n^2+(11/2)n-(1/2)(n-1)^2-(11/2)(n-1)
=n+5
A1=S1=1/2+11/2=6，也满足上式
An=n+5

B(n+2)+Bn=2B(n+1)
{Bn}是等差数列
B1=B3-2d=11-2d
S9=9B1+(1/2)×9×8×d
=9(11-2d)+36d
=18d+99=153
d=3
B1=11-2×3=5
Bn=B1+(n-1)d=5+(n-1)3=3n+2

2.
Cn=6/[(2(An)-11)(2(Bn)-1)]
=6/[(2(n+5)-11)(2(3n+2)-1)]
=6/[(2n-1)(6n+3)]
=2/[(2n-1)(2n+1)]
=[(2n+1)-(2n-1)]/[(2n-1)(2n+1)]
=1/(2n-1)-1/(2n+1)
Tn=C1+C2+C3+……+Cn
=(1/1-1/3)+(1/3-1/5)+(1/5-1/7)+……+(1/(2n-1)-1/(2n+1))
=1-1/(2n+1)
当n=1时，Tn取最小值
T1=1-1/3=2/3>k/57
38>k
使上式成立的最大正整数k=37