由a1=1,a(n+1)=2an+1,可得,a1=1,a2=3,a3=7,a4=15.又a(n+1)+1=2[an+1].===>[a(n+1)+1]/(an+1)=2.∴{an+1}是首项为2,公比为2的等比数列.∴an+1=2^n.∴通项为an=2^n-1.(n=1,2,3,).