设数列{an}的前n项的和为Sn,且方程x^2-anx-an=0,有一根为Sn-1,n=1,2,3,...,求a1、a2

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设数列{an}的前n项的和为Sn,且方程x^2-anx-an=0,有一根为Sn-1,n=1,2,3,...,求a1、a2.

[s(n)-1]^2 -a(n)[s(n)-1] - a(n) = 0,
n=1时,a(1)=s(1),
0 = [s(1)-1]^2 - a(1)[s(1)-1] - a(1) = [a(1)-1]^2 - a(1)[a(1)-1] - a(1) = [a(1)-1][a(1)-1-a(1)] - a(1)
= [a(1)-1](-1) - a(1)
= 1 - 2a(1),
a(1)=1/2.
n=2时,s(2)=a(1)+a(2)=a(2)+1/2,
0 = [s(2)-1]^2 - a(2)[s(2)-1] - a(2) = [s(2)-1][s(2)-1-a(2)] - a(2)
= [a(2)+1/2-1][a(2)+1/2-1-a(2)] - a(2)
= [a(2)-1/2](-1/2) - a(2)
= 1/4 - 3a(2)/2,
a(2) = 1/6
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