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

来源：学生作业帮助网编辑：作业帮时间：2024/11/16 15:57:28

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

设f（x)满足f(-x)=-f(x),f(x)=f(4-x),x∈[0,2)时,f(x)=x,则f(11.5)等于? f(x+1)+f(x-1)=2x²-4x,求f（x）? f(x+1)+f(x-1)=4x^3-2x求f(x) f(x)=min{x^2-2,x,4-x} 怎样由f(x+2)=1/f(x)得出f(x+4)=f(x) f(x+2)=-f(x)怎么得到f(x+4)=f(x)? f(x)= 4-x^2 ,|f(x)|2求f(f(x)) 若函数f(x)满足f(x)+2f(1/x)=5x+4/x,则f(x)= 二次函数f(x)满足f(x+1)+f(x-1)=2x^2+4x,求f(x) 函数f(x)={(1/2)^x,(x>4),f(x+3),(x 函数f(x)={(1/2)^x,(x>4),f(x+3),(x 已知f(x)={2^x,(x≥4);f(x+1)(x 设函数f(x)={(1/2)^x(x≥4),f(x+3)(x f(x)=x+4,x f(x)=x+4,x 函数f(x)=2-x,(x>=0);f(x)=|x^2+4x+2|(x 数学f(x)=x(x+1)(x+2)(x+3)(x+4)(x+5),求f'(0)=? f(x)=x(x-1)(x+2)(x-3)(x+4)……(x+100),求f'(1)