an+a(n+1)=-3nan*a(n+1)=cn+9*n^2/4a2=-3-a1=-4[an+(3/2)n]+[a(n+1)+(3/2)(n+1)]=3/2设bn=an+(3/2)n,b1=a1+(3/2)=5/2,b2=a2+3=-1bn+b(n+1)=3/2b(n-1)+bn=3/2两式相减：b(n+1)-b(n-1)=0所以错项相等,邻项之和为3/2b(2k-1)=b1=5/2b(2k)=b2=-1a(2k-1)+(3/2)(2k-1)=b(2k-1)=5/2a(2k)+(3/2)(2k)=b(2k)=-1a(2k-1)=-(3/2)(2k-1)+5/2=-3k+4a(2k)=-(3/2)(2k)-1=-3k-1 当n=2k-1时,an*a(n+1)=a(2k-1)*a(2k)=(-3k+4)(-3k-1)=9k^2-9k-4=c(2k-1)+9*(2k-1)^2/4c(2k-1)=-9*(2k-1)^2/4+9k^2-9k-4=-25/4当n=2k时,an*a(n+1)=a(2k)*a(2k+1)=(-3k-1)(-3k+1)=9k^2-1=c(2k)+9*(2k)^2/4=c(2k)+9*k^2c(2k)=-9*k^2+9k^2-1=-1 所以c1+c3+……+c(2k-1)+……+c1999=1000(-25/4)=-6250c2+c4+……+c(2k)+……+c2000=1000(-1)=-1000所以c1+c2+c3+…+c2000=[c1+c3+……+c(2k-1)+……+c1999]+[c2+c4+……+c(2k)+……+c2000]=-6250-1000=-7250
我来试试看x^2+3nx+cn+9*n^2/4=0 由韦达定理得，它的两解x1,x2x1x2=9n^2/4 x1+x2=3n则ana(n+1)=9n^2/4+cn an+a(n+1)=3n 脑子不灵敏了，明天再告诉你
