对于正数X,规定f(X)=x/1+X,例如:f(3)=3/(1+3)=3/4,f(1/3)=(1/3)/(1+(1/3))=1/4,计算f(1/2006)+...+f(1/3)+f(1/2)+f(1)+f(2)+f(3)+...+f(2006)=()
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![对于正数X,规定f(X)=x/1+X,例如:f(3)=3/(1+3)=3/4,f(1/3)=(1/3)/(1+(1/3))=1/4,计算f(1/2006)+...+f(1/3)+f(1/2)+f(1)+f(2)+f(3)+...+f(2006)=()](/uploads/image/z/6917554-10-4.jpg?t=%E5%AF%B9%E4%BA%8E%E6%AD%A3%E6%95%B0X%2C%E8%A7%84%E5%AE%9Af%28X%29%3Dx%2F1%2BX%2C%E4%BE%8B%E5%A6%82%EF%BC%9Af%283%29%3D3%2F%281%2B3%29%3D3%2F4%2Cf%281%2F3%29%3D%281%2F3%29%2F%281%2B%281%2F3%29%29%3D1%2F4%2C%E8%AE%A1%E7%AE%97f%281%2F2006%29%2B...%2Bf%281%2F3%29%2Bf%281%2F2%29%2Bf%281%29%2Bf%282%29%2Bf%283%29%2B...%2Bf%282006%29%3D%EF%BC%88%EF%BC%89)
对于正数X,规定f(X)=x/1+X,例如:f(3)=3/(1+3)=3/4,f(1/3)=(1/3)/(1+(1/3))=1/4,计算f(1/2006)+...+f(1/3)+f(1/2)+f(1)+f(2)+f(3)+...+f(2006)=()
对于正数X,规定f(X)=x/1+X,例如:f(3)=3/(1+3)=3/4,f(1/3)=(1/3)/(1+(1/3))=1/4,计算f(1/2006)+...+f(1/3)+f(1/2)+f(1)+f(2)+f(3)+...+f(2006)=()
对于正数X,规定f(X)=x/1+X,例如:f(3)=3/(1+3)=3/4,f(1/3)=(1/3)/(1+(1/3))=1/4,计算f(1/2006)+...+f(1/3)+f(1/2)+f(1)+f(2)+f(3)+...+f(2006)=()
f(x)+f(1/x)=x/(1+x)+[(1/x)/(1+(1/x))]=x/(1+x)+1/(1+x)=1
所以:
原式=[f(1/2006)+f(2006)]+[f(1/2005)+f(2005)]+[f(1/2004)+f(2004)]+...+[f(1/2)+f(2)]+f(1)
=2005+f(1)
=2005.5
分析f(3)=3/(1+3)=3/4,f(1/3)=(1/3)/(1+(1/3))=1/4,似乎有规律,所以
f(X)=x/(1+x);所以f(1/x)=(1/x)/(1+1/x)=1/(1+x),所以f(X)+f(1/x)=1
所以f(1/2006)+...+f(1/3)+f(1/2)+f(1)+f(2)+f(3)+...+f(2006)=2006-f(1)=2006-0.5=2005.5
当x=1时,f(1)=12;
当x=2时,f(2)=13,当x=12时,f(12)=23;
当x=3时,f(3)=14,当x=13时,f(13)=34…,
故f(2)+f(12)=1,f(3)+f(13)=1,…,所以f(n)+…+f(1)+…+f(1n)=f(1)+(n-1),由此规律即可得出结论.
∵当x=1时,f(1)=12,当x=2时,f(2)=13,当x=...
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当x=1时,f(1)=12;
当x=2时,f(2)=13,当x=12时,f(12)=23;
当x=3时,f(3)=14,当x=13时,f(13)=34…,
故f(2)+f(12)=1,f(3)+f(13)=1,…,所以f(n)+…+f(1)+…+f(1n)=f(1)+(n-1),由此规律即可得出结论.
∵当x=1时,f(1)=12,当x=2时,f(2)=13,当x=12时,f(12)=23;当x=3时,f(3)=14,当x=13时,f(13)=34…,
∴f(2)+f(12)=1,f(3)+f(13)=1,…,
∴f(n)+…+f(1)+…+f(1n)=f(1)+(n-1),
∴f(2012)+f(2011)+…+f(2)+f(1)+f(12)+…+f(12011)+f(12012)=f(1)+(2012-1)=12+2011=2011.5.
故答案为:2011.5.分析f(3)=3/(1+3)=3/4,f(1/3)=(1/3)/(1+(1/3))=1/4,似乎有规律,所以
f(X)=x/(1+x);所以f(1/x)=(1/x)/(1+1/x)=1/(1+x),所以f(X)+f(1/x)=1
所以f(1/2006)+...+f(1/3)+f(1/2)+f(1)+f(2)+f(3)+...+f(2006)=2006-f(1)=2006-0.5=2005.5
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