1/1*3+1/3*5+1/5*7+...+1/99*101=?(要简便)

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1/1*3+1/3*5+1/5*7+...+1/99*101=?(要简便)
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1/1*3+1/3*5+1/5*7+...+1/99*101=?(要简便)
1/1*3+1/3*5+1/5*7+...+1/99*101=?(要简便)

1/1*3+1/3*5+1/5*7+...+1/99*101=?(要简便)
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
=(1-1/101)÷2
=100/101÷2
=50/101

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

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