已知各项均为正数的数列{an}的前n项和Sn满足6Sn=(an+1)(an+2),n∈N+,且a

n=1时,6a1=a12+3a1+2,且a1＞1,解得a1=2．
n≥2时,6Sn=an2+3an+2,6Sn-1=an-12+3an-1+2,两式相减得（an+an-1）（an-an-1-3）=0,
∵an+an-1＞0,
∴an-an-1=3,
∴{an}为等差数列,
∵a1=2,
∴an=3n-1．
（II）证明：∵数列{bn}满足an(2bn−1)＝1,
∴bn＝log2
3n
3n−1
∴Tn=b1+b2+…+bn=log2(
3
2
×
6
5
×…×
3n
3n−1
)
要证2Tn+1＜log2（an+3）,即证2log2(
3
2
×
6
5
×…×
3n
3n−1
)+1＜log2（an+3）
即证(
3
2
×
6
5
×…×
3n
3n−1
)2＜
3n+2
2
即证
2(
3
2
×
6
5
×…×
3n
3n−1
)2
3n+2
＜1
令cn＝
2(
3
2
×
6
5
×…×
3n
3n−1
)2
3n+2
,
∴
cn+1
cn
＝
9n2+18n+9
9n2+21n+10
＜1
∵cn＞0,∴cn+1＜cn,
∴{cn}是单调递减数列
∴cn≤c1＝
2×(
3
2
)2
3×1+2
＝
9
10
＜1
∴cn＝
2(
3
2
×
6
5
×…×
3n
3n−1
)2
3n+2
＜1
故2Tn+1＜log2（an+3）．