数列{an}中,a1=1,an>0且对所有n恒有S(n+1)+Sn=[S(n+1)-Sn]^2,求an,Sn_数学解答

数列{an}中,a1=1,an>0且对所有n恒有S(n+1)+Sn=[S(n+1)-Sn]^2,求an,Sn

人气：235 ℃　时间：2020-05-31 08:19:30

解答

S(n+1)+Sn=[S(n+1)-Sn]^2=a(n+1)^2
Sn+S(n-1)=[Sn-S(n-1)]^2=an^2
[S(n+1)+Sn]-[Sn+S(n-1)]=a(n+1)+an=a(n+1)^2-an^2
则a(n+1)-an=1
an=a1+(n-1)=n,Sn=(a1+an)*n/2=n*(n+1)/2