若a+b=1,ab=-1,h1=a+b,h2=a^2+b^2,h3=a^3+b^3,h4=a^4+b^4

题目

若a+b=1,ab=-1,h1=a+b,h2=a^2+b^2,h3=a^3+b^3,h4=a^4+b^4
(1)求h1*h2*h3*h4的值
(2)h7的值

答案

a+b=1
ab=-1
h1=a+b=1
h2=a^2+b^2=(a+b)^2-2ab=1^2-2*(-1)=3
h3=a^3+b^3=(a+b)(a^2-ab+b^2)=1*[(a+b)^2-3ab]=1^2-3*(-1)=4
h4=a^4+b^4=(a^2+b^2)^2-2(ab)^2=3^2-2*(-1)^2=9-2=7
h1h2h3h4=1*3*4*7=84
h7=a^7+b^7=(a^3+b^3)(a^4+b^4)-a^3b^4-a^4b^3
= (a^3+b^3)(a^4+b^4)-a^3b^3(a+b)
= h3h4-(ab)^3(a+b)
= 4*7 - (-1)^3 * 1
= 28 + 1
= 29