题目

设数列[an}的前n项和为Sn,已知a1=1且满足3Sn的平方=an(3Sn-1),1求{1/Sn}为等差数列 1.求{1/Sn}为等差数列
2.若bn=Sn/3n+1,数列{bn}的前n项和为Tn,求Tn

提问时间：2020-10-09

答案

1.n>1,3Sn^2=An(3Sn -1)An=Sn-S(n-1)下标3Sn^2=(3Sn -1)[Sn-S(n-1)]=3Sn^2-3SnS(n-1)-Sn+S(n-1)3SnS(n-1)=S(n-1)-Sn[S(n-1)-Sn]/[SnS(n-1)]=31/Sn - 1/S(n-1) = 31/Sn=1+3(n-1)=3n-2 当n=1时也满足.所以{1/Sn}为等差...