高一化简题目

f(x+y)=f(x)+f(y)
f(x+0)=f(x)+f(0)
f(0)=0
f(x+(-x))=f(x)+f(-x)=0
f(-x)=-f(x)
f(2x)=f(x+x)=2f(x)
f(x)=(1/2)f(2x)
f(1/2)=(1/2)f(1)
同理：f(1/3)=(1/3)f(1)
...
f(1/n)=(1/n)f(1）

f(n/(n+1))=f(1-(1/(n+1)))=f(1)+f(-1/(n+1))=f(1)-(1/(n+1))f(1)=(1-(1/(n+1)))f(1)
所以：
f(1/2)+f(2/3)+f(3/4)+......f(2008/2009)
=f(1)*(2008-((1/2)+(1/3)+(1/4)+...+(1/2009))

f(1/2)+f(2/3)+f(3/4)+......f(2008/2009)=f（1/2+2/3+……+2008/2009)=f（1-1/2+1-1/3+1-1/4+…+1-1/2009)=f[2008-(1/2+1/3+……+1/2009)]