(Ⅰ)由a n+1=a n 2+6a n+6得a n+1+3=(a n+3) 2,
∴
log ( a n+1 +3)5 =2
log ( a n +3)5 ,即c n+1=2c n
∴{c n}是以2为公比的等比数列.
(Ⅱ)又c 1=log 55=1,
∴c n=2 n-1,即
log ( a n +3)5 =2 n-1,
∴a n+3= 5 2 n-1
故a n= 5 2 n-1 -3
(Ⅲ)∵b n=
1
a n -6 -
1
a n 2 +6 a n =
1
a n -6 -
1
a n+1 -6 ,∴T n=
1
a 1 -6 -
1
a n+1 -6 =-
1
4 -
1
5 2 n -9 .
又0<
1
5 2 n -9 ≤
1
5 2 -9 =
1
16 .
∴-
5
16 ≤T n<-
1
4