设f（x）=4^x/(4^x +2),求和S=f(1/2002)+f(2/2002)+……+f(2001/2002)

f(1-x)=4^(1-x)/[4^(1-x)+2]上下乘4^x,4^(1-x)*4^x=4^(1-x+x)=4所以f(1-x)=4/(4+2*4^x)=2/(4^x+2)所以f(x)+f(1-x)=(4^x+2)/(4^x+2)=1所以原式=[f(1/2002)+f(2001/2002)]+……+[f(1000/2002)+f(1002/2002)]+f(1001/2002)=[f(1/2002)+f(1-1/2002)]+……+[f(1000/2002)+f(1-1000/2002)]+[f(1001/2002)+f(1001/2002)]/2=1+1+……+1+1/2=1000+1/2=2001/2=