已知a+b=1,ab=-1,设s1=a+b,s2=a^2+b^2,s3=a^3+b^3.Sn=a^n+b^n,(1)求S2,S3,S4的值(2)试写出Sn,Sn+1,Sn+2三者之间的关系(3)根据以上得出的结论,计算a^8+b^8
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1.S1=a+b=1
S2=a^2+b^2=(a+b)^2-2ab=3
S3=a^3+b^3=(a+b)^3-3ab(a+b)=4
S4=a^4+b^4=(a^2+b^2)^2-2(ab)^2=7
2.Sn-2+Sn-1=Sn
3.S8=S7+S6
=S5+S6+S4+S5
=S4+S3+S4+S4+S3+S4+S4+S3
=5S4+3S3
=35+12=47