a(n+1)+an=4n-3,an+a(n-1)=4*(n-1)-3,
故a(n+1)-a(n-1）=4,（n≥2)
a1=2,a2=-1
當(dāng)n為奇數(shù)時(shí),
an=2+(n-1）/2*4=2n,a(n-1)=-1+(n-1)/2*4=2n-5,
故Sn=(2+2n)*(n+1)/2/2+(-1+2n-5)*(n-1)/2/2
=n^2-n+2
當(dāng)n為偶數(shù)時(shí),
an=-1+2n,a(n-1)=2+2n,
故Sn=(2+2+2n)*n/2/2+(-1+2n-1)*n/2/2=n^2+n/2