Question

在数列{an}中a1=0,an+1+an=n平方+2n求通项公式

Answer 1

a(n+1)+an=n^2+2n
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 =2
-2k2+1=2
k2= -1/2

coef. of constant
-2k3-k1-k2=0
-2k3+1/2+1/2=0
k3= 1/2

ie
a(n+1) -(1/2)(n+1)^2-(1/2)(n+1) +1/2 = -( an -(1/2)n^2-(1/2)n+1/2 )
=>{an -(1/2)n^2-(1/2)n+1/2} 是等比数列, q=-1
an -(1/2)n^2-(1/2)n+1/2 = (-1)^(n-1) .(a1 -1/2-1/2+1/2)
=(-1)^n .(1/2)
an =(1/2)n^2+(1/2)n-1/2 +(-1)^n .(1/2)

Answer 2

通向公式是什么，解释？