已知数列{an}中.a1=1,n大于等于2时.其前n项和Sn满足Sn^2=an(Sn-1/2).求证：数列{1/Sn}是等差数列.

由Sn^2=an(Sn-1/2),两边同时除以Sn,拆开括号,得到Sn=an-an/2Sn,移项,an-Sn=an/2Sn,两边同时除以an,乘以2,得到2(an-Sn)/an=1/Sn,
那么1/(Sn-1)=2[(an-1)-(Sn-1)]/(an-1),第一个减第二个,得到1/Sn-1/(Sn-1)=2｛an(Sn-1)-(an-1)Sn}/an(an-1),将Sn=(Sn-1 )+an代入,替换Sn,化简得到1/Sn-1/(Sn-1)=2,所以是公差为2的等差数列.
n和n-1项有点乱.看小心点吧o(︶︿︶)o设bn=Sn/2n+1.求数列{bn}的前n项和Tn