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
问题描述:
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()
答
y=kx+k-1当y=00=kx+k-1x=(1-k)/ky=(k+1)x+k当y=00=(k+1)x+kx=-k/(k+1)y=kx+k-1=(k+1)x+kx=-1y=-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×...