已知数列{an}的前n项和为Sn,且有Sn=n^2/2+11n/2,数列｛bn｝满足bn+2-2bn+1+bn=（n∈N+）,且b3=11,前9项和为153

1、
a1=S1=1/2+11/2=6
Sn=n²/2 +11n/2
S(n-1)=(n-1)²/2 +11(n-1)/2
S(n-1)-Sn=an=n²/2 +11n/2 -(n-1)²/2 -11(n-1)/2=n+5
n=1时,a1=1+5=6,同样满足.
数列{an}的通项公式为an=n+5.
b(n+2)-2b(n+1)+bn=0
b(n+2)-b(n+1)=b(n+1)-bn
数列{bn}是等差数列,设公差为d.
S9'=9b1+36d=9(b1+4d)=9b5=153
b5=17
b5-b3=2d=17-11=6d=3
b1=b3-2d=11-6=5
bn=5+3(n-1)=3n+2
数列{bn}的通项公式为bn=3n+2.
2、
cn=3/[(2an-11)(2bn-1)]=2/[2(n+5)-11][2(3n+2)-1]=2/[(2n-1)(6n+3)]
=(2/3)/[(2n-1)(2n+1)]=(1/3)[1/(2n-1)-1/(2n+1)]
Tn=c1+c2+...+cn
=(1/3)[1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)]
=(1/3)[1-1/(2n+1)]
=2n/[3×(2n+1)]
=2/[3×(2+ 1/n)]
随n增大,1/n减小,2+ 1/n减小,3×(2+ 1/n)减小,2/[3×(2+ 1/n)]增大,当n=1时,Tn取得最小值Tmin=2/[3×(2+1)]=2/9
Tn>k/57对于一切n∈N+都成立,则当Tn取得最小值时等式同样成立.
2/9>k/57
k