已知数列{an}a1=1 n>=2 Sn^2=an(Sn-1/2)证明是{1/Sn}等差数列

显示全部楼层 · 2009-7-6 19:08:54

麻烦大家

千问 · 2009-7-6 19:08:54

Sn^2=an[Sn-1/2]因为an=Sn-S(n-1)代入上式得到Sn^2=(Sn-S(n-1))[Sn-1/2]整理S(n-1)-Sn=2SnS(n-1)左右同时除以SnS(n-1)得到1/Sn-1/S(n-1)=2所以{1/Sn}是等差数列，公差为2，首项为1/S1=1...

千问 · 2009-7-6 19:08:54

an=Sn-S(n-1)所以(Sn)^2=[Sn-S(n-1)](Sn-1/2)=(Sn)^2-1/2*Sn-Sn*S(n-1)+1/2*S(n-1)-1/2*Sn-Sn*S(n-1)+1/2*S(n-1)=0S(n-1)-Sn=2Sn*S(n-1)[S(n-1)-Sn]/Sn*S(n-1)=2S(n-1)/Sn*S(n-1)-Sn/Sn...