Question

设数列{an}的前n项和为Sn,且方程x^2-anx-an=0有一个根为Sn-1,求an的通项公式

Answer 1

(Sn-1)^2-(Sn-1)an-an=0

an=(Sn-1)^2/Sn

又an=Sn-S(n-1)

Sn-S(n-1)=(Sn-1)^2/Sn

Sn^2-SnS(n-1)=Sn^2-2Sn+1

Sn=1/(2-S(n-1))
两边同减1，取倒数

1/(Sn-1)=1/(1/(2-S(n-1))-1)=(2-S(n-1))/(S(n-1)-1)=1/(S(n-1)-1)-1

所以{1/(Sn-1)}是等差数列

1/(Sn-1)=1/(S1-1)+(n-1)(-1)=-n-1

Sn=n/(n+1)

an=Sn-S(n-1)=n/(n+1)-(n-1)/n=1/n(n+1)