数列{an}中,a1=1,an+1=-an+n^2,求{an}的通项及a2000.求详解,

数列{an}中,a1=1,an+1=-an+n^2,求{an}的通项及a2000.求详解,

a(n+1)=-an+n^2
let
a(n+1) +k1(n+1)^2+k2(n+1) + k3 = -( an +k1n^2+k2n+k3)
coef.of n^2
-2k1=1
k1 = -1/2
coef.of n
-2k2-2k1=0
-2k2+1=0
k2 =1/2
coef.of constant
-2k3 -k1-k2 =0
-2k3+1/2-1/2=0
k3=0
ie
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
( an -(1/2)n^2+(1/2)n)/( a1 -(1/2)+(1/2)) = (-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
再问： 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？~
再答： 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
再问： 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！~
再答： 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+k3 bn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