设直线y=－n／n+1X+√2／n+1（n为正整数）与两坐标轴围成的三角形面积为sn(n=1,2,……,2009),则s1+s2+s3+……+s2009的值为多少?

Question

设直线y=－n／n+1X+√2／n+1（n为正整数）与两坐标轴围成的三角形面积为sn(n=1,2,……,2009),则s1+s2+s3+……+s2009的值为多少?

Answer 1

当X=0时 Y=√2/(n+1)
当Y=0时
nx/(n+1)=√2/(n+1)
x=√2/n
Sn=1/2 *x *y=1/2 *√2/n*√2/(n+1)=1/n(n+1)
所以S1+S2+S3+.S2009
=1/1*2+1/2*3+1/3*4+.1/2009*2010
=1-1/2 +1/2-1/3+1/3-1/4+.1/2009-1/2010
=1-1/2010
=2009/2010