f(x)=9^x/(9^x+3),求f(1/2000)+f(2/2000)+……+f(1999/2000)
问题描述：
f(x)=9^x/(9^x+3),求f(1/2000)+f(2/2000)+……+f(1999/2000)
答
∵f(x)+f(1-x)=9^x/(9^x+3)+9^(1-x)/(9^(1-x)+3)=(9+3*9^x+9+3*9^(1-x))/(9+9+3*9^x+3*9^(1-x))=1∴原式=[f(1/2000)+f(1999/2000)]+[f(2/2000)+f(1998/2000)]+……+f(1000/2000)=999+f(1/2)=999+3/(3+3)=999.5