當前位置:編程學習大全網 - 行動軟體 - 數列An=x^n+y^n+z^n的遞推式是怎麽證明的。。。。An+3=(x+y+z)An+2-(xy+yz+xz)An+1+xyzAn。給完答案後加分

數列An=x^n+y^n+z^n的遞推式是怎麽證明的。。。。An+3=(x+y+z)An+2-(xy+yz+xz)An+1+xyzAn。給完答案後加分

這題妳沒給出A1A2A3的初值,我就根據妳給出的通項設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)

設An+1-yAn=bn b1=x^2+z^2-xy-yz b2=x^3+z^3-x^2y-z^2y

那麽就是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)

還可得bn+2-zbn+1=x(bn+1-zbn),那麽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)]

設An-(z^n+x^n)=Cn,那麽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

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