（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 ．
