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.
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.