Question

设直线Y=KX+K-1和直线Y=（K+1）X+K(K是正整数)与X轴围成的三角形面积为Sk,则S

S1+S2+S3+....+S2012的值为（）

Answer 1

y=kx+k-1
当y=0
0=kx+k-1
x=(1-k)/k
y=(k+1)x+k
当y=0
0=(k+1)x+k
x=-k/(k+1)
y=kx+k-1=(k+1)x+k
x=-1
y=-1
面积=|(1-k)/k+k/(k+1)|×|-1|÷2
=1/2|k(k+1)|
因为 Sk k=1,2,3,......
所以面积=1/2k(k+1)
S1+S2+S3+.....+S2012
=(1/2)(1/2+1/2×3+1/3×4+1/4×5+1/5×6+.....+1/2012×2013)
=(1/2)(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+.....-1/2012+1/2012-1/2013)
=(1/2)(1-1/2013)
=1/2×2012/2013
=1006/2013

Answer 2

K=1
Y1=x
K=2012, Y2 = 2013x + 2012
只需计算y1与Y2和X轴之间的面积