歸納法問題??

Let f(x) = x^( 2n-1 ) + y^( 2n-1 )

By Factor Theorem, if f( -y ) = 0, f(x) is divisible by ( x + y ).

http://en.wikipedia.org/wiki/Factor_theorem
( A deeper approach to Factor Theorem )

Put x = -y into x^( 2n-1 ) + y^( 2n-1 )
f(x) = ( -y )^( 2n - 1 ) + y^( 2n - 1 )

No matter n is even or odd, 2n is an even number
So, ( 2n - 1 ) must be an odd number
So f(x) = - y^( 2n - 1 ) + y^( 2n - 1 )
= 0

As f( -y) = 0, f(x) is divisible by ( x + y )

不用歸納法就能證明了

x^(2n+1)+y^(2n+1)
=(x + y)(x^2n - x^(2n-1) y + x^(2n-2)y^2 - ... + y^2n)

[最後一項是加的原因是因為它的x的指數是0，是一個雙數。]

要知道為甚麼，乘開就知道了。

(x + y)(x^2n - x^(2n-1) y + x^(2n-2)y^2 - ... + y^2n)
= (x^(2n+1) - x^(2n) y + x^(2n-1)y^2 - ... + xy^2n) +
(x^(2n)y - x^(2n-1) y^2 + x^(2n-2)y^3 - ... + y^(2n+1))

留意，很巧的，中間的項都全消了，我是用長除法知道這答案的。

使用2n + 1 而不使用2n - 1 是表達上的方便，都是單數嘛，沒關係的。

2007-05-18 22:00:27 補充：
你快左七分鐘 :'(