已知数列﹛an﹜的前n项和为Sn,a1=1,且2nSn+1－2(n+1)Sn=n²+n(n∈N*).（1）求数列{an}的通项公式；（2）设bn=n/2(n+3)Sn,求数列{bn}的前n项和Tn；（3）证明：n≥2时,1/（a2的三次方）+1/（a3的三次方）+1/(a4的三次方）+……+1/（an的三次方）＜1/4