Question

设数列{an}的前n项和为Sn,已知a1=1,且3Sn^2=an(3Sn-1),n大于等于2.

Answer 1

（1）3Sn^2=An（3Sn-1）,即

3Sn^=[Sn-S(n-1)]（3Sn-1）=3Sn^2-Sn-3SnS(n-1)+S(n-1),即

1/Sn-1/S(n-1)=3

所以{1/Sn}是等差数列

(2)

1/Sn=1/s1+d(n-1)=1/A1+3(n-1)=3n-2

Sn=1/(3n-2)谢谢给个好评