(1)数列{an}的前n项之和Sn=(-1) n(2n 2+4n+1)-1,在n=1时,a 1=s 1=(-1) 1(2+4+1)-1=-8
在n≥2时,a n=s n-s n-1=(-1) n(2n 2+4n+1)-(-1) n-1[2(n-1) 2+4(n-1)+1]=(-1) n•4n(n+1),
而n=1时,a 1=-8满足a n=(-1) n4n(n+1),故所求数列{a n}通项a n=(-1) n4n(n+1).
(2)∵b n=
(-1) n
a n =
1
4n(n+1) =
1
4 (
1
n -
1
n+1 ),
因此数列{b n}的前n项和T n=
1
4 (1-
1
n+1 )=
4n
n+1