∵数列{an}的各项均为正数,
∴an+1+an>0,
∴an+1-2an=0,
即an+1=2an,所以数列{an}是以2为公比的等比数列.
∵a3+2是a2,a4的等差中项,
∴a2+a4=2a3+4,
∴2a1+8a1=8a1+4,
∴a1=2,
∴数列{an}的通项公式an=2n.
∵数列{bn}满足b1=1,且bn+1=bn+2.
∴{bn}是首项为1,公差为2的等差数列,
∴bn=1+(n-1)×2=2n-1.…(6分)
(2)∵cn=
1?(?1)n |
2 |
1+(?1)n |
2 |
=
1?(?1)n |
2 |
1+(?1)n |
2 |
=
|
∴T2n=(2+23+…+2n-1)-[(3+7+…+(4n-1)]
=
2(1?4n) |
1?4 |
n(3+4n?1) |
2 |
=
22n+1?2 |
3 |