(I)证明:∵an=2an-1+2n-1(n≥2),∴an−1=2(an−1−1)+2n,
∴
an−1
2n=
an−1−1
2n−1+1.∴bn=bn-1+1.
∴{bn}是首项为
a1−1
2=
5−1
2=2,公差为1的等差数列;
(II)由(I)可得bn=2+(n-1)×1=n+1,
∴
an−1
2n=n+1,∴an=(n+1)•2n+1,
令cn=(n+1)•2n,其前n项和为Tn,
则Tn=2×2+3×22+4×23+…+n•2n-1+(n+1)•2n,
2Tn=2×22+3×23+…+n•2n+(n+1)•2n+1,
两式相减得-Tn=2×2+22+23+…+2n-(n+1)•2n+1=2+
2(2n−1)
2−1-(n+1)•2n+1=-n•2n+1,
∴Tn=n•2n+1.
∴Sn=Tn+n=n+n•2n+1.