Question

一条直线kx+(k+1)y-1=0与坐标轴围成的三角形的面积为Sk,求S1+S2+S3+S4+...S2009的值

Answer 1

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=2009代入

上式=2009/（2*2009+2）=2009/4020