∵an=3an-1+2n-1(n≥2),
∴an+1=3an+2n+1,
两式相减得an+1-an=3an-3an-1+2,
整理得:an+1-an+1=3(an-an-1+1),
又∵a1=4,a2=3a1+3=15,
∴a2-a1+1=15-4+1=12,
∴an+1-an+1
=3•(an-an-1+1)
=32•(an-1-an-2+1)
=…
=3n-1•(a2-a1+1)
=4•3n,
∴an+1=(an+1-an+1)+(an-an-1+1)+…+(a2-a1+1)+a1-n
=4(3n+3n-1+…+3)+4-n
=4•3(1-3