Y=x^2/(1+x^2),=f(x)则f(1)+f(2)+f(3)+...+f(n)+f(1/2)+f(1/3)+...+f(1/n)=_____
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![Y=x^2/(1+x^2),=f(x)则f(1)+f(2)+f(3)+...+f(n)+f(1/2)+f(1/3)+...+f(1/n)=_____](/uploads/image/z/12550662-54-2.jpg?t=Y%3Dx%5E2%2F%281%2Bx%5E2%29%2C%3Df%28x%29%E5%88%99f%281%29%2Bf%282%29%2Bf%283%29%2B...%2Bf%28n%29%2Bf%281%2F2%29%2Bf%281%2F3%29%2B...%2Bf%281%2Fn%29%3D_____)
Y=x^2/(1+x^2),=f(x)则f(1)+f(2)+f(3)+...+f(n)+f(1/2)+f(1/3)+...+f(1/n)=_____
Y=x^2/(1+x^2),=f(x)则f(1)+f(2)+f(3)+...+f(n)+f(1/2)+f(1/3)+...+f(1/n)=_____
Y=x^2/(1+x^2),=f(x)则f(1)+f(2)+f(3)+...+f(n)+f(1/2)+f(1/3)+...+f(1/n)=_____
f(1/x)=(1/x)^2/[1+(1/x)^2]上下乘x^2
=1/(1+x^2)
所以f(x)+f(1/x)=x^2/(1+x^2)+1/(1+x^2)=(1+x^2)/(1+x^2)=1
所以f(1)=1/(1+1)=1/2
f(2)+f(1/2)=1
……
f(n)+f(1/n)=1
所以f(1)+f(2)+f(1/2)+f(3)+f(1/3)+……+f(n)+f(1/n)
=1/2+1+1+……+1
=1/2+(n-1)
=n-1/2
这个题目,要先证明
f(x)+f(1/x)=1
这个很容易证明啊
因此
f(1)+f(2)+f(3)+...+f(n)+f(1/2)+f(1/3)+...+f(1/n)
=f(1)+[f(2)+f(1/2)]+[f(3)++f(1/3)]...+[f(n)+f(1/n)]
=1/2+n-1
=n-1/2
f(1/x)=1/(x²+1)
f(x)+f(1/x)=1
(1)+f(2)+f(3)+...+f(n)+f(1/2)+f(1/3)+...+f(1/n)=n-1/2
因为f(n)=n^2/(1+n^2)=1-1/(1+n^2)
而f(1/n)=(1/n)^2/[1+(1/n)^2]=1/(1+n^2)
所以f(n)+f(1/n)=1
所以f(1)+f(2)+....+f(n)+f(1/2)+f(1/3)+....+f(1/n)=f(1)+(n-1)=1/2+n-1=n-1/2