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,
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

你这个数列有问题，如果n=1第二个式子是不成立的