Question

2.设直线L1:y=kx+k-1和直线L2:y=(k+1)x+k (k为正整数)及 x轴围成的三角形面积为Sk,

2.设直线L1:y=kx+k-1和直线L2:y=(k+1)x+k
(k为正整数)及 x轴围成的三角形面积为Sk,
求S1+S2+S3+…….+S2011的值。

Answer 1

L1:y=kx+k-1与x轴交点：0=kx+k-1，x1=-(k-1)/k
L2:y=(k+1)x+k与x轴交点：0=(k+1)x+k，x2=-k/(k+1)
k为正整数，x1＞x2
L1与L2交点横坐标：kx+k-1=(k+1)x+k，x3=-1
y3=-k+k-1=-1
三角形面积S(k)=1/2(x1-x2)*|y3|=1/2[-(k-1)/k+k/(k+1)]=1/[2k(k+1)]

S1+S2+S3+…….+S2011
=1/[2*1*2]+1/[2*2*3]+1/[2*3*4]+......+1/[2*2011*2012]
=1/2 {1/[1*2]+1/[2*3]+1/[3*4]+......+1/[2011*2012] }
=1/2 {1-1/2+1/2-1/3+1/3-1/4+......+1/2011-1/2012 }
=1/2 (1-1/2012)
=1/2 * 2011/2012
=2011/4024

Answer 2

交点（(1-k)/k,0),(-k/(k+1),0),(-1,-1)
SK=1/2*|(1-k)/k+k/(1+k)]*1
=1/21/[k(k+1)]
=1/2*[1/k-1/(k+1)]
S1+S2+S3+…….+S2011
=1/2(1-1/2+1/3-1/3+...-1/2010+1/2010-1/2011)
=1/2(1-1/2011)
=1005/2011