数列{an}中,a1=1sn是{an}的前n项和,当n大于等于2是sn=an[1-2/sn]求证{1/sn}是等差数列

Sn = an ( 1 - 2 / Sn ) = ( Sn - S(n-1)) ( 1 - 2 / Sn )
= Sn - S(n-1) - 2 + 2 S(n-1) / Sn
因为 S1 = a1 = 1 ,当 n > 2 时,Sn != 0 ,
所以 0 = - 1 - 2 / S(n-1) + 2 / Sn ,
所以 1 / Sn - 1 / S(n-1) = 1/2 ,n = 2 ,3 ,4 …
所以 {1/Sn}是等差数列