举报
旧dd见青山
a(n+1)=-an+n^2 设 a(n+1) +k1(n+1)^2+k2(n+1) + k3 = -( an +k1n^2+k2n+k3) n^2 的系数 -2k1=1 k1 = -1/2 n的系数 -2k2-2k1=0 -2k2+1=0 k2 =1/2 常数的系数 -2k3 -k1-k2 =0 -2k3+1/2-1/2=0 k3=0 故得 a(n+1) -(1/2)(n+1)^2+(1/2)(n+1) = -( an -(1/2)n^2+(1/2)n) [a(n+1) -(1/2)(n+1)^2+(1/2)(n+1)]/( an -(1/2)n^2+(1/2)n) =-1 设 bn= an -(1/2)n^2+(1/2)n bn 是等比数列, q=-1 b(n+1)/bn =-1 bn/b1= (-1)^(n-1) bn =(-1)^(n-1) an -(1/2)n^2+(1/2)n = (-1)^(n-1) an = (1/2)n^2 -(1/2)n + (-1)^(n-1) a2000 = 1998999