f(x)=(4^x)/(4^x+2),那么和式f(1/1001)+f(2/1001)+f(3/1001).+f(1000/1001)的值为多少.
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![f(x)=(4^x)/(4^x+2),那么和式f(1/1001)+f(2/1001)+f(3/1001).+f(1000/1001)的值为多少.](/uploads/image/z/8952557-5-7.jpg?t=f%EF%BC%88x%EF%BC%89%3D%EF%BC%884%5Ex%EF%BC%89%2F%284%5Ex%2B2%29%2C%E9%82%A3%E4%B9%88%E5%92%8C%E5%BC%8Ff%281%2F1001%29%2Bf%282%2F1001%29%2Bf%283%2F1001%EF%BC%89.%2Bf%281000%2F1001%29%E7%9A%84%E5%80%BC%E4%B8%BA%E5%A4%9A%E5%B0%91.)
f(x)=(4^x)/(4^x+2),那么和式f(1/1001)+f(2/1001)+f(3/1001).+f(1000/1001)的值为多少.
f(x)=(4^x)/(4^x+2),那么和式f(1/1001)+f(2/1001)+f(3/1001).
+f(1000/1001)的值为多少.
f(x)=(4^x)/(4^x+2),那么和式f(1/1001)+f(2/1001)+f(3/1001).+f(1000/1001)的值为多少.
f(x)=(4^x)/(4^x+2)=1-2/(4^x+2)
f(1-x)=1-2/[4^(1-x)+2]=1-4^x/(4^x+2)
所以:f(x)+f(1-x)==1-2/(4^x+2)+1-4^x/(4^x+2)
=2-1=1
所以f(1/1001)+f(2/1001)+f(3/1001).
+f(1000/1001)=f(1/1001)+f(1000/1001)
+f(2/1001)++f(999/1001)
f(3/1001)+.+ f(500/1001)+f(501/1001)=500×1=500
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(2/100...
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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
收起
因为f(x)+f(1-x)=1
下面的自己应该可以自己算了吧.