Question

已知正数数列{an}的前n项和为Sn,且对所有的正整数n,an与2的等差中项等于Sn与2的等比中项，求an

Answer 1

(an+2)/2=√(2*Sn) => Sn=1/8an^2+1/2an+1/2 --(1)
S(n-1)=1/8a(n-1)^2+1/2a(n-1)+1/2 --(2)
(1)-(2) an=1/8(an^2-a(n-1)^2)+1/2(an-a(n-1))
化简: 1/8(an^2-a(n-1)^2)-1/2(an+a(n-1))=0
=> (an+a(n-1))(an-a(n-1)-4)=0 (3)
因为an>0 an+a(n-1)≠0， ：. an-a(n-1)-4 => an-a(n-1)=4
为等差数列。
令n=1,由（1）得 a1=1/8a1^2+1/2a1+1/2 => a1=2
则 an=a1+(n-1)d=2+(n-1)*2=2n