f(x)=(4^x)/((4^x)+2).求f（1/1001）+f(2/1001)+…+f(1000/1001)的值

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f(x)=(4^x)/((4^x)+2).求f（1/1001）+f(2/1001)+…+f(1000/1001)的值
f(x)=(4^x)/((4^x)+2).求f（1/1001）+f(2/1001)+…+f(1000/1001)的值

f(x)=(4^x)/((4^x)+2).求f（1/1001）+f(2/1001)+…+f(1000/1001)的值
f(x)+f(1-x)
=4^x/(4^x+2)+4^(1-x)/[4^(1-x)+2]
=4^x/(4^x+2)+(4/4^x)/[(4/4^x)+2]
=4^x/(4^x+2)+4/(4+2*4^x)
=4^x/(4^x+2)+2/(2+4^x)
=(4^x+2)/(4^x+2)
=1
所以f(1/1001)+f(2/1001)+……+f(1000/1001)
=[f(1/1001)+f(1000/1001)]+……+[f(500/1001)+f(501/1001)]
=1+1+……+1
=1*500
=500

500

f(1-x)=4^(1-x)/[4^(1-x)+2]
=4*4^(-x)/[4*4^(-x)+2]
上下同乘以4^x
f(1-x)=4/(4+2*4^x)=2/(2+4^x)
所以f(x)+f(1-x)=(2+4^x)/(2+4^x)=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*1
=500

易知f(x)+f(1-x)=1
然后把要求的式子乘以2，首位相加，然后再除2

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