已知数列{an},满足;a1=a(0<a<1),an+1+an=(-1)^n,Sn为数列{an}前n项和

a(2)+a(1)=(-1)^1*1=-1, a(2)=-a-1. a(2n+1)+a(2n)=(-1)^(2n)*(2n)=2n, a(2n-1+1)+a(2n-1)=(-1)^(2n-1)*(2n-1)=1-2n, (2n)-(1-2n) = a(2n+1)+a(2n)-[a(2n)+a(2n-1)]=a(2n+1)-a(2n-1)=4n-1, a(2n+1)=a(2n-1)+4n-1=a(2n-1)+2[n(n+1)-(n-1)n] + n - (n+1), a(2n+1)-2n(n+