已知数列an中,a1=1,当n大于等于2时,sn=an(1-2/sn).求证1/sn是等差数列

n>1时sn=an(1-2/sn)=(sn-s(n-1))(1-2/sn)=sn-s(n-1)-2+2s(n-1)/sn整理可得：sn*s(n-1)=2(s(n-1)-sn)1/sn-1/s(n-1)=(s(n-1)-sn)/sn*s(n-1)=1/2故1/sn是公差为1/2的等差数列