已知:f(x)=4^x/(4^x+2)
=2^2x/(2^2x+2)=2^/[2^+1]………………………(1)
那么,f(1-x)=2^/[2^+1]
=2^/[2^+1](分子分母乘以2^得到,)
=2/[2+2^]
=1/[2^+1]…………………………………………………(2)
(1)+(2)得到:
f(x)+f(1-x)=[2^+1]/[2^+1]=1
所以:
f(1/1001)+f(2/1001)+...+f(1000/1001)
=[f(1/1001)+f(1000/1001)]+[f(2/1001)+f(999/1001)]+……+[f(499/1001)+f(502/1001)]+[f(500/1001)+f(501/1001)]
=1+1+……+1(一共500个)
=500
f(1-x)+f(x) =4^(1-x)/[4^(1-x)+2] +4^x/(4^x+2) =1
=== f(1/1001)+f(2/1001)+...+f(1000/1001)
= [f(1/1001)+f(1000/1001)]+[f(2/1001)+f(999/1001)]+....+
+ [f(500/1001)+f(501/1001)]
= 500
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