100-(1+ $\frac{1}{2}$ + $\frac{1}{3}$ +…..+ $\frac{1}{100}$ )/ $\frac{1}{2}$ + $\frac{2}{3}$ + $\frac{3}{4}$+……+ $\frac{99}{100}$
100-(1+ $\frac{1}{2}$ + $\frac{1}{3}$ +…..+ $\frac{1}{100}$ )/ $\frac{1}{2}$ + $\frac{2}{3}$ + $\frac{3}{4}$+……+ $\frac{99}{100}$
$\frac{100-(1+1/2+1/3+…+1/100)}{1/2+2/3+3/4+…+99/100}$=$\frac{(1-1)+(1-1/2)+(1-1/3)+…+(1-1/100)}{1/2+2/3+3/4+…+99/100}$ =$\frac{1/2+2/3+3/4+…+99/100}{1/2+2/3+3/4+…+99/100}$ =1
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