(1)∵数列{an}是等差数列,a1=1,a2=2,∴an=n.再由数列{bn}满足bn=an+1+(-1)nan,n∈N+.
可得 b1=b3=b5=1,b2=5,b4=9,b6=13,∴数列{bn}的前6项和S6=30.
(2)∵数列{bn}是公差为2的等差数列,b1=a2-a1=1,∴bn =2n-1.
再由bn=an+1+(-1)nan可得b2n-1=a2n-a2n-1=4n-3,b2n=a2n+1+a2n=4n-1.
相减可得 a2n+1+a2n-1=2,a2n+3+a2n+1=2,∴a2n+3=a2n-1.
∵a1=1,a3=1,∴a4n-3=a1=1,a4n-1=a3=1.
∴an=
.
1 , n为奇数 2n-2 , n为偶数