a(n+1)=-an+n^2leta(n+1) +k1(n+1)^2+k2(n+1) + k3 = -( an +k1n^2+k2n+k3)coef.of n^2-2k1=1k1 = -1/2coef.of n-2k2-2k1=0-2k2+1=0k2 =1/2coef.of constant-2k3 -k1-k2 =0-2k3+1/2-1/2=0k3=0iea(n+1) -(1/2)(n+1)^2...谢谢，不过你写的那些符号是什么意思唉，我看不懂额，我是个高中生，麻烦能不能解释一下？~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=1k1 = -1/2n的系数-2k2-2k1=0-2k2+1=0k2 =1/2常数的系数-2k3 -k1-k2 =0-2k3+1/2-1/2=0k3=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)nbn 是等比数列， q=-1b(n+1)/bn =-1bn/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设a(n+1) +k1(n+1)^2+k2(n+1) + k3 = -( an +k1n^2+k2n+k3)这一步是什么意思，哪儿得到的？是固定用法吗？我就这里看不懂额，麻烦你解释一下，谢谢！~把原有的a(n+1)=-an+n^2变成这样a(n+1) +k1(n+1)^2+k2(n+1) + k3 = -(an +k1n^2+k2n+k3)那bn=an +k1n^2+k2n+k3bn就是一个等比数列然后根据a(n+1)=-an+n^2再计算出k1, k2,k3 的值