f( 1-x)= 2( 1-x)/[ 1-2( 1-x)]=(2x-2)/( 1-2x)；

∴f(x)+f( 1-x)=-2；

Sn=f( 1/n)+f(2/n)+f(3/n)+....+f(n- 1/n) ①

sn = f((n- 1)/n)+f((n-2)/n)+……+f( 1/n)②

From ①+②:

2Sn =[f( 1/n)+f((n- 1)/n)]+[f(2/n)+f((n-2)/n)]+……+[f((n- 1)/n)+f( 1/n)]；

When n is an even number, there must be f((n/2)/n)=f( 1/2), which is meaningless.

2Sn=-2*(n-2)

Sn=2-n

When n is odd, f((n/2)/n)=f( 1/2) does not exist.

2Sn=-2*(n- 1)

Sn= 1-n

To sum up:

When n is an even number: Sn=2-n

When n is odd, sn =1-n.