Question

1/1×3+1/3×5+1/5×7+……+1/99×100 简算

Answer 1

解：
利用性质 1/[n(n+2)]=(1/2)[1/n-1/(n+2)]
题目有误，前后不匹配
∴ 1/1×3+1/3×5+1/5×7+……+1/99×101
=(1/2)(1-1/3+1/3-1/5+1/5-1/7+......+1/99-1/101)
=(1/2)(1-1/101)
=(1/2)*(100/101)
=50/101

Answer 2

因为1/1×3=(1-1/3)×1/2 1/3×5=(1/3-1/5)×1/2 1/5×7=(1/5-1/7)×1/2 ………
所以原式=(1-1/3+1/3-1/5+1/5-1/7+……+1/99-1/101)×1/2
=(1-1/101) ×1/2
=50/101

Answer 3

原式=1/2（1-1/3+1/3-1/5+1/5-1/7+……+1/99-1/101)=50/101

Answer 4

裂项相消，若是三项，N项亦可

Answer 5

=1/2*(1-1/3+1/3-1/5+1/5-1/7+.....+1/99-1/101)
=1/2*(1-1/101)
=50/101