证明
bn+1-bn=Tn+1-Tn-[(n+2)平方-(n+1)平方]/2
=1/(an+1*an+2)-1/(a1*a2)-[(n+2)平方-(n+1)平方]/2 (Tn+1-Tn按照前n项和展开可以得到）
=根号【（n+1)(n+2)】-(2n+3)/2
=根号【（n+1)(n+2)】-根号【（n+3/2)(n+3/2)】
=根号【n^2+3n+2】-根号【n^2+3n+9/4】