Question

已知数列（an)中,an>0,Sn=a1+a2+...+an,且2根号Sn=1+an,求an

Answer 1

2根号Sn=1+An

4Sn=(1+An)^2

4A1=4S1=(1+A1)^2

(A1-1)^2=0

A1=1

n>=2时

4An=4Sn-4S(n-1)=(1+An)^2-(1+A(n-1))^2

(1+A(n-1))^2=(1+An)^2-4An=(An-1)^2

An-1=±(1+A(n-1))

如果An-1=-(1+A(n-1))

An=-A(n-1)

A2=-A1=-1