数列An=x^n+y^n+z^n的递推式是怎么证明的.An+3=(x+y+z)An+2-(xy+yz+xz)An+1+xyzAn.以免被吞An+3=(x+y+z)An+2-(xy+yz+xz)An+1+xyzAn

Question

数列An=x^n+y^n+z^n的递推式是怎么证明的.An+3=(x+y+z)An+2-(xy+yz+xz)An+1+xyzAn.
以免被吞
An+3=(x+y+z)An+2 -(xy+yz+xz)An+1 +xyzAn

Answer 1

这题你没给出A1A2A3的初值,我就根据你给出的通项设A1=x+y+z A2=x^2+y^2+z^2
An+3=(x+y+z)An+2 -(xy+yz+xz)An+1 +xyzAn
An+3-yAn+2=(x+z)An+2-y(x+z)An+1-xzAn+1+xyzAn
(An+3-yAn+2)=(x+z)(An+2-yAn+1)-xz(An+1-yAn)
设An+1-yAn=bn b1=x^2+z^2-xy-yz b2=x^3+z^3-x^2y-z^2y
那么就是bn+2=(x+z)bn+1-xzbn
可得bn+2-xbn+1=z(bn+1-xbn),所以bn+1-xbn=(b2-xb1)*z^(n-1)=(z-x)(z-y)z^n （1）
还可得bn+2-zbn+1=x(bn+1-zbn),那么bn+1-zbn=(b2-zb1)*x^(n-1)=(z-x)(y-x)x^n （2）
z*(1)-x*(2)得到：bn+1=(z-y)*z^(n+1)-(y-x)*x^(n+1)
也就是An+1-yAn=(z-y)*z^n-(y-x)*x^n=z^(n+1)+x^(n+1)-y(z^n+x^n)
所以An+1-[z^(n+1)+x^(n+1)]=y[An-(z^n+x^n)]
设An-(z^n+x^n)=Cn,那么Cn+1=yCn,所以Cn等比
Cn=C1*y^n-1=y^n
所以An-(x^n+z^n)=y^n
所以An=x^n+y^n+z^n