(1)∵an=an-1+2n,∴an-1=an-2+2(n-1),…,a2=a1+2?2将这n-1个式子相加得:an=n(n+1)…(4分)(2)∵ 1 an = 1 n(n+1) = 1 n ? 1 n+1 …(6分)∴Sn=1? 1 2 + 1 2 ? 1 3 +…+ 1 n ? 1 n+1 ,∴Sn=1? 1 n+1 .…(8分)∵n+1≥2,∴0< 1 n+1 ≤ 1 2 ,∴ 1 2 ≤Sn<1.…(12分)