若f(x)=(x-1/2)^3,则f(0.1)+f(0.2)+f(0.3)+...+f(0.9)=____

f（x）=-0.4^3-0.3^3-0.2^3-0.1^3+0+0.1^3+0.2^3+0.3^3+0.4^3=0

f(1-x)=-(x-1/2)^3
f(0.1)+f(0.2)+f(0.3)+...+f(0.9)
=f(0.1)+f(0.9)+f(0.2)+f(0.8)+f(0.3)+f(0.7)+f(0.4)+f(0.6)+f(0.5)
=0+0+0+0+0
=0

f（0.1）= - f（0.9），
f（0.2）= - f（0.8），
f（0.3）= - f（0.7），
f（0.4）= - f（0.6），
所以f（0.1）+f（0.2）+……+f（0.9）=f（0.5）=0

f(0.1)+f(0.2)+f(0.3)+...+f(0.9)=（-0.4）^3+（-0.3）^3+（-0.2）^3=-0.099

f(0.1)=(0.1-1/2)^3=-(0.9-1/2)^3=-f(0.9)
f(0.2)=(0.2-1/2)^3=-(0.8-1/2)^3=-f(0.8)
f(0.3)=(0.3-1/2)^3=-(0.7-1/2)^3=-f(0.7)
f(0.4)=(0.4-1/2)^3=-(0.6-1/2)^3=-f(0.6)
f(0.5)=(0.5-1/2)^3=0
所以,f(0.1)+f(0.2)+f(0.3)+...+f(0.9)=0.
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