an+1=an/(1+2an) => 1/a(n+1)=(1+2an)/an=1/an+2∴1/a(n+1)-1/an=2∴{1/an}是等差数列由1/a(n+1)-1/an=2可得1/an-1/a(n-1)=21/a(n-1)-1/a(n-2)=2......1/a2-1/a1=2将上述a2到an共n-1个等式加起来,得1/an-1/a1=2*(n-1)1/an=1/a1+2(n-1)=1+2n-2=2n-1∴an=1/(2n-1)