奧數 如何解答

As a^2 = a + 1 and b^2 = b + 1,
so, a and b are the roots of the equation x^2 - x - 1 = 0.
ie. a + b = 1, ab = -1, therefore
a^5 + b^5
= (a + b)(a^4 - a^3b + a^2b^2 - ab^3 + b^4)
= (a + b)[(a + b)^4 - 5(a^3b + ab^3) - 5a^2b^2]
= (a + b){(a + b)^4 - 5ab[(a + b)^2 - 2ab] - 5(ab)^2}
= (1){1^4 - 5(-1)[1^2 - 2(-1)] - 5(-1)^2}
= 1 + 5(3) - 5
= 11

a，b為x^2－x－1=0之兩根
a+b=1，ab=－1
x^2=x+1
a^2=a+1
b^2=b+1
a^2+b^2=(a+b)+2=3
x^3=x^2+x
a^3=a^2+a
b^3=b^2+b
a^3+b^3=(a^2+b^2)+(a+b)=3+1=4
x^4=x^3+x^2
a^4=a^3+a^2
b^4=b^3+b^2
a^4+b^4=(a^3+b^3)+(a^2+b^2)=4+3=7
x^5=x^4+x^3
a^5=a^4+a^3
b^5=b^4+b^3
a^5+b^5=(a^4+b^4)+(a^3+b^3)=7+4=11

讚﹗根本5系常人諗到ge野……

那些年是正確的～

但如果不想處理高次數的算式，可以由以上的繼續

　5a + 5b + 6
= 5(a+b) + 6
= 5(1) + 6
= 11

a*a = a + 1
a*a*a = (a + 1)*a = a*a + a = (a + 1) + a = 2a + 1
a*a*a*a = (2a + 1)*a = 2a*a + a = 2(a + 1) + a = 3a + 2
a*a*a*a*a = (3a + 2)*a = 3a*a + 2a = 3(a + 1) + 2a = 5a + 3

Similarly, b*b*b*b*b = 5b + 3

　a*a*a*a*a + b*b*b*b*b

= (5a + 3) + (5b + 3)

= 5a + 5b + 6

2013-10-17 22:48:49 補充：
那些年是正確的～

但如果不想處理高次數的算式，可以由以上的繼續

　5a + 5b + 6
= 5(a+b) + 6
= 5(1) + 6
= 11