请教高手一道数学问题

设a+b=1  ,a^2+b^2=2,求a^7+b^7=?

要有完整的解题过程。帮帮忙

(a+b)^2=1
a^2+b^2+2ab=1
2+2ab=1
ab=-0.5

(a^2+b^2)^2=4
a^4+b^4+2(ab)^2=4
a^4+b^4=3.5
(a^4+b^4)^2=12.25
a^8+b^8+2(ab)^4=12.25
a^8+b^8=12.125

a^6+b^6
=(a^2+b^2)[a^4+b^4-(ab)^2]
=2*(3.5-0.25)
=6.5

a^7+b^7
=(a^7+b^7)*(a+b)
=a^8+b^8+ab^7+ba^7
=(a^8+b^8)+ab(a^6+b^6)
=12.125+(-0.5)*6.5
=8.875

采纳下哈   谢谢