三數之和 - 拼圖
By Faithe
at 2019-12-22T20:50
at 2019-12-22T20:50
Table of Contents
: x + y + z = 1
: x^2 + y^2 + z^2 = 2
: x^3 + y^3 + z^3 = 3
: 求
: x^5 + y^5 + z^5 = ?
: 延伸問題:x^n+y^n+z^n = ?
: 提示:請別強行去解x,y,z,會少許多解題樂趣!
三個未知數 三個方程式
先求出它們的關係吧
x+y+z=1
xy+yz+zx=?
xyz=?
(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)
=>1=2+2(xy+yz+zx)
=>xy+yz+zx=-(1/2)
x^3+y^3+z^3-3xyz=(x+y+z)[x^2+y^2+z^2-(xy+yz+zx)]
=>3-3xyz=1[2+(1/2)]
=>xyz=1/6
不會化簡x^5+y^5+z^5 只好拜託估狗 找到了這個
(x+y+z)^5-x^5-y^5-z^5=5(y+z)(x+y)(z+x)(x^2+y^2+z^2+xy+yz+zx)
又x+y+z=1
所以
(x+y+z)^5-x^5-y^5-z^5=5(1-x)(1-y)(1-z)(x^2+y^2+z^2+xy+yz+zx)
=5[1-(x+y+z)+(xy+yz+zx)-xyz](x^2+y^2+z^2+xy+yz+zx)
=>1-(x^5+y^5+z^5)=5[1-1-(1/2)-(1/6)][2-(1/2)]
=>x^5+y^5+z^5=6
順便算了一下x^4+y^4+z^4=25/6
可能還要多算幾項才能找到x^n+y^n+z^n的一般式吧 我放棄 O_Q
--
: x^2 + y^2 + z^2 = 2
: x^3 + y^3 + z^3 = 3
: 求
: x^5 + y^5 + z^5 = ?
: 延伸問題:x^n+y^n+z^n = ?
: 提示:請別強行去解x,y,z,會少許多解題樂趣!
三個未知數 三個方程式
先求出它們的關係吧
x+y+z=1
xy+yz+zx=?
xyz=?
(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)
=>1=2+2(xy+yz+zx)
=>xy+yz+zx=-(1/2)
x^3+y^3+z^3-3xyz=(x+y+z)[x^2+y^2+z^2-(xy+yz+zx)]
=>3-3xyz=1[2+(1/2)]
=>xyz=1/6
不會化簡x^5+y^5+z^5 只好拜託估狗 找到了這個
(x+y+z)^5-x^5-y^5-z^5=5(y+z)(x+y)(z+x)(x^2+y^2+z^2+xy+yz+zx)
又x+y+z=1
所以
(x+y+z)^5-x^5-y^5-z^5=5(1-x)(1-y)(1-z)(x^2+y^2+z^2+xy+yz+zx)
=5[1-(x+y+z)+(xy+yz+zx)-xyz](x^2+y^2+z^2+xy+yz+zx)
=>1-(x^5+y^5+z^5)=5[1-1-(1/2)-(1/6)][2-(1/2)]
=>x^5+y^5+z^5=6
順便算了一下x^4+y^4+z^4=25/6
可能還要多算幾項才能找到x^n+y^n+z^n的一般式吧 我放棄 O_Q
--
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