Maths~~巧解難題

a+b+c=1
a*a+b*b+c*c=2
a*a*a+b*b*b+c*c*c=3
問a*a*a*a+b*b*b*b+c*c*c*c=?

a+b+c=1
a*a+b*b+c*c=2
(a+b+c)(a*+b*+c*)=2
1x(a*+b*+c*)=2
所以(a*+b*+c*)=2

a*a*a+b*b*b+c*c*c=3
(a+b+c)(a*a*+b*b*+c*c*)=3
1x(a*a*+b*b*+c*c*)=3
所以(a*a*+b*b*+c*c*)=3

a*a*a*a+b*b*b*b+c*c*c*c=?
(a+b+c)(a*a*a*+b*b*b*+c*c*c*)=?
1x(a*a*a*+b*b*b*+c*c*c*)=?
(a*a*a*+b*b*b*+c*c*c*)=?
a*既3次+b*既3次+c*既3次
所以a*a*a*a+b*b*b*b+c*c*c*c=a*^3+b*^3+c*^3

但係照讀又應該=4啵 =00=
哈哈~

已知a + b + c = 1，a2 + b2 + c2 = 2，a3 + b3 + c3 = 3，

問a4 + b4 + c4 = ?


這時候你需要用Newton' s formulas for the sum of powers of the roots（根之冪和的牛頓公式）。

甚麼是Newton' s formulas for the sum of powers of the roots（根之冪和的牛頓公式）？

Let f(x) ≡ a0xn + a1xn - 1 + …… + an be a

polynomial of degree n and let α1, α2, ……, αn

be the n roots of the equation f(x) = 0 . For each

positive integer r, we define

Sr = Σ(i = 1 to n) αir and also define S0 = n ,


then a0Sr + a1Sr - 1 + a2Sr - 2 + …… + ar - 1S1 + rar = 0

for 1 ≦ r ≦ n - 1

a0Sr + a1Sr - 1 + a2Sr - 2 + …… + anSr - n = 0

for r ≧ n


回到你的問題：


This is no need to find a, b and c separately by solving

the given equation. a, b and c can be considered as the

roots of the cubic equation a0x3 + a1x2 + a2x + a3 = 0 ,

where a0, a1, a2 and a3 are constants.

If Sr = ar + br + cr , Then S1 = 1 , S2 = 2 , S3 = 3 .

Let S0 = 3 ,

by Newton' s formulas, we have

a0S1 + 1a1 = 0

a0 + a1 = 0

其實a0 let甚麼數都可以（除了0之外），

不過我們為了方便計算，就let a0 = 1:

1 + a1 = 0

a1 = - 1


a0S2 + a1S1 + 2a2 = 0

1 × 2 + (- 1) × 1 + 2a2 = 0

1 + 2a2 = 0

a2 = - 1/2


a0S3 + a1S2 + a2S1 + a3S0 = 0

1 × 3 + (- 1) × 2 + (- 1/2) × 1 + a3 × 3 = 0

1/2 + 3a3 = 0

a3 = - 1/6


a0S4 + a1S3 + a2S2 + a3S1 = 0

1 × S4 + (- 1) × 3 + (- 1/2) × 2 + (- 1/6) × 1 = 0

S4 - 25/6 = 0

S4 = 25/6


∴a4 + b4 + c4 = 25/6

你的問題:

已知a+b+c=1,a*a+b*b+c*c=2,a*a*a+b*b*b+c*c*c=3,問a*a*a*a+b*b*b*b+c*c*c*c=?

我的答案:

i think it is:

a=1/6

b=1/3

c=1/2

------so,a+b+c= 1/6+1/3+1/2=1

a+a+b+b+c+c=1/6+1/6+1/3+1/3+1/2+1/2=2

a+a+a+b+b+b+c+c+c=1/6+1/6+1/6+1/3+1/3+1/3+1/2+1/2+1/2=3

so,

a+a+a+a+b+b+b+b+c+c+c+c=1/6+1/6+1/6+1/6+1/3+1/3+1/3+1/3+1/2+1/2+1/2+1/2=4

ok??

2007-11-01 16:46:03 補充：
sor~ 我錯了,a+b+c=1a*a+b*b+c*c=2(a+b+c)(a*b*c)=21x(a*b*c)=2so,(a*b*c)=2a*a*a+b*b*b+c*c*c=3(a+b+c)(a*a*+b*b*+c*c)=31x(a*a*+b*b*+c*c)=3so,(a*a*+b*b*+c*c)=3

2007-11-01 16:46:52 補充：
a*a*a*a+b*b*b*b+c*c*c*c=?=(a+b+c)(a*a*a*+b*b*b*+c*c*c)=?=1x(a*a*a*+b*b*b*+c*c*c)=?=(a*a*a*+b*b*b*+c*c*c)=?=a^3*b^3*c^3so,a*a*a*a+b*b*b*b+c*c*c*c=a*^3+b*^3+c*^3but , if it is an iq question,the answer may be four, yet?