kx+(k+1)y-1=0
当x=0时y=1/(k+1)
当y=0时x=1/k
所以S=1/2*1/(k+1)*1/k=1/[2k*(k+1)]
所以S1=1/(2*1*2)
S2=1/(2*2*3)
S3=1/(2*3*4)
.
Sk=1/[2*k*(k+1)]
S和=1/(2*1*2)+1/(2*2*3)+1/(2*3*4)+.+1/[2*k*(k+1)]
=1/2*{1/(1*2)+1/(2*3)+1/(3*4)+.+1/[K*(K+1)]}
=1/2*{1/1-1/2+1/2-1/3+1/3-1/4+.+1/k-1/(k+1)
=1/2*{1-1/(k+1)}
=k/(2k+2)
把k=2004代入
上式=2004/（2*2004+2）=1002/2005