已知a+b=1,a^2+b^2=2,求 a^7+b^7.

a+b=1,a^2+b^2=2
(a+b)^2=a^2+b^2+2ab=1
2ab=-1
ab=-1/2
a,b
联想韦达定理
是
xx－x－1/2＝0的2根
xx＝x＋1/2
aa=a+1/2
bb=b+1/2
a^n=a^(n-1)+(1/2)a^(n-2)
b^n=b^(n-1)+(1/2)b^(n-2)
S(n)=a^n+b^n
=S(n-1)+(1/2)S(n-2)
S1=1
S2=-1/2
S3=S1+1/2 S2=3/4
S4=S2+1/2 S3=-1/2+3/8=-1/8
S5=S4+1/2 S3
=-1/8+3/8=1/4
S6=S5+1/2 S4
=1/4+1/2(-1/8)
=3/16
S7=S6+1/2(S5)
=3/16+1/2(1/4)
=5/16

a+b=1,a^2+b^2=2
(a+b)^2=a^2+b^2-2ab
1=2-2ab
ab=0.5
a^3+b^3=(a+b)^3-3ab(a+b)=1-3*0.5*1=-0.5
a^4+b^4=(a^2+b^2)^2-2(ab)^2=4-2*(0.5)^2=3.5

a^7+b^7
=(a^3+b^3)*(a^4+b^4)-(a^3b^4+a^4b^3)
=(-0.5)*3.5-(ab)^3*(a+b)
=-1.75-(0.5)^3*1
=-1.875

a^3+b^3=(a+b)(a^2+b^2-ab)

a^5+b^5=(a+b)^5-5ab[2ab(a+b)+a^3+b^3]

a^7+b^7=(a+b)^7-7ab[3ab(a^3+b^3)+(a^5+b^5)+5a^2b^2(a+b)]

由于
a+b=1
a^2+b^2=2=(a+b)^2-2ab

所以
2=