Question

设直线kx+（k+1）y-1=0与坐标轴所围成的直角三角形的面积为Sk,求S1+S2+.+S2012.

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=2012代入

上式=2012/（2*2012+2）=1006/2013