(1)利用an+1=an+2n,求出a2=a1+2,a3=a2+22,a4=a3+23,an=an-1+2n-1(n≥2),然后各式相加即可求出{an}的通项公式.
(2)由bn=n(1+an),求出bn=n•2n,Sn=2+2•22+3•23++n•2n和2Sn=22+2•23++(n-1)•2n+n•2n+1,两式相减即可求出Sn.
【解析】
(1)∵a2=a1+2,a3=a2+22,a4=a3+23,
an=an-1+2n-1(n≥2)
相加,得an=a1+2+22+…+2n-1=2n-1,
又a1=1符合上式
∴an=2n-1,
(2)bn=n•2n,Sn=2+2•22+3•23++n•2n,
2Sn=22+2•23++(n-1)•2n+n•2n+1,
∴Sn=-(2+22+23++2n)+n•2n+1=2-2n+1+n•2n+1 .