(1)设数列{an}的公差为d,数列{bn}的公比为q,
则an=1+(n-1)d,bn=2qn?1,
由b1b3=b4,得q=
=b1=2,b4 b3
∴an=2n-1,bn=2n.
(2)T2n+1=c1+a1+(a2+b1)+a3+(a4+2?b2)+…+a2n-1+(a2n+nbn)
=1+S2n+(b1+2b2+…+nbn),
令A=b1+2b2+…+nbn,
则A=2+2?22+…+n?2n,
2A=22+2?23+…+(n-1)?2n+n?2n+1,
∴-A=2+22+…+2n-n?2n+1,
∴A=?
+n?2n+1?2n+1+2,2(1?2n) 1?2
∵S2n=
=4n2,2n(1+a2n) 2
∴T2n+1=1+4n2+n?2n+1?2n+1+2
=3+4n2+(n-1)?2n+1.