1/1*3+1/3*5+1/5*7+...+1/99*101=?(要简便)

利用公式1/(n)(n+2)=1/2[1/n-1/(n+2)]
1/3*1+1/3*5+1/5*7+...+1/99*101
=1/2（1-1/3）+1/2（1/3-1/5）……1/2(1/99-1/101)
=1/2(1-1/3+1/3-1/5……1/99-1/101)
=1/2（1-1/101）
=1/2*100/101
=55/101

1/1×3+1/3×5+1/5×7+...+1/99×101
=(1/2)×(2/1×3+2/3×5+2/5×7+...+2/99×101)
=(1/2)×[(1-1/3)+(1/3-1/5)+(1/5-1/7)+...+(1/99-1/101)]
=(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

1/1*3+1/3*5+1/5*7+...+1/99*101=(1-1/3)÷2+(1/3-1/5)÷2+(1/5-1/7)÷2+……+(1/99-1/101)÷2=[(1-1/3)+(1/3-1/5)+(1/5-1/7)+……+(1/99-1/101)]÷2=(1-1/3+1/3-1/5+1/5-1/7+……+1/99-1/101)÷2中间互相抵消=(1-1/...