解:
(1)
设公差为d。数列单调递增,则d>0
a2、a3、a6成等比数列,则a3²=a2·a6
(a2+d)²=a2·(a2+4d)
d²-2a2·d=0
d(d-2a2)=0
d=0(舍去)或d=2a2
a2=1,d=2a2=2
a1=a2-d=1-2=-1
an=a1+(n-1)d=-1+2(n-1)=2n-3
数列{an}的通项公式为an=2n-3
(2)
Sn=(a1+an)n/2=(-1+2n-3)n/2=n(n-2)
bn=1/S(n+2)=1/[n(n+2)]=½[1/n -1/(n+2)]
Tn=b1+b2+...+bn
=½[1/1-⅓+½-¼+...+1/n-1/(n+2)]
=½[1+½ -1/(n+1)-1/(n+2)]
=¾ -1/(2n+2) -1/(2n+4)