数列an的前n项的和sn,当n≥1时,s(n+1)是a(n+1)与s(n+1)+2的等比中项

问题描述:

数列an的前n项的和sn,当n≥1时,s(n+1)是a(n+1)与s(n+1)+2的等比中项
求证:数列{1/sn}为等差数列
当a1=-1时,求an

,s(n+1)是a(n+1)与s(n+1)+2的等比中项 ==》》a(n+1)*(s(n+1)+2)=s(n+1)^2a(n+1)=s(n+1)-sn(s(n+1)-sn)*(s(n+1)+2)=s(n+1)^22s(n+1)-sn*s(n+1)-2sn=01/s(n+1)-1/sn=-1/2 数列{1/sn}为等差数列 1/sn=1/s1-1/2(n-...