(1)a2=2a1=2,a3=a2+1=3,
∵a2n+1=a2n+1=2a2n-1+1,
∴a2n+1+1=2(a2n-1+1),
∴数列{a2n-1+1}为公比是2的等比数列;
(2)S2n+1=(a1+a2)+(a3+a4)+…+(a2n-1+a2n)+a2n+1+a2n+1
=3a1+3a3+…+3a2n-1+a2n+1
由(1)知,
∴a2n?1+1=2n,
∴a2n?1=2n?1
∴S2n+1=3[(2?1)+(22?1)+…+(2n?1)]+a2n+1=3(2
?n)+2n+1?1=2n+3-3n-71?2n
1?2