当x1+x2=1
则:f(x2)=4^(1-x1)/[4^(1-x1)+2]=4/[4+2*4^x1]=2/[4^x1+2]
所以:f(x1)+f(x2)=4^x1/[4^x1+2] +2/[4^x1+2]=1
而:k1/2009 + k2008/2009=(1+2008)/2009=1
所以:f(k1/2009)+f(k2008/2009)=1
同理:f(k2/2009)+f(k2007/2009)=1
f(k3/2009)+f(k2006/2009)=1
.
所以:
s=f(k1/2009)+f(k2/2009)+.+f(k2008/2009)
=[f(k1/2009)+f(k2008/2009)] + [f(k2/2009)+f(k2007/2009)] + [f(k3/2009)+f(k2006/2009)] + .
=1+1+1+...
=2008/2
=1004