1\(1乘3)+1/(3乘5)+1/(5乘7)+...+1/(99乘101)

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1\(1乘3)+1/(3乘5)+1/(5乘7)+...+1/(99乘101)1\(1乘3)+1/(3乘5)+1/(5乘7)+...+1/(99乘101)1\(1乘3)+1/(3乘5)+1/(5乘7)+

1\(1乘3)+1/(3乘5)+1/(5乘7)+...+1/(99乘101)
1\(1乘3)+1/(3乘5)+1/(5乘7)+...+1/(99乘101)

1\(1乘3)+1/(3乘5)+1/(5乘7)+...+1/(99乘101)
因为1/(1*3)=1/2*(1/1-1/3)
1/(3*5)=1/2*(1/3-1/5)
1/(5*7)=1/2*(1/5-1/7)
同理推导可得
1/(99*101)=1/2*(1/99-1/101)
所以原式=1/2*(1/1-1/3+1/3-1/5+1/5-1/7+.+1/97-1/99+1/99-1/101)
=1/2*(1/1-1/101)
=50/101

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

2/1*3+2/3*5+2/5*7+......+2/99*101
=1-1/3+1/3-1/5+1/5-1/7+......+1/99-1/101
=1-1/101
=100/101

1/(1*3)+1/(3*5)+1/(5*7)+……+1/(99*101)
=2[(1-1/3)+(1/3-1/5+(1/5-1/7)+……(1/99-1/101)]
=2[1-1/101]=200/101