Question

Calculați:(1/100+2/101+3/102+.......+101/200-101):(1/100+1/101+.....+1/200)
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Answer 1

$\dfrac{\frac{1}{100}+\frac{2}{101}+\frac{3}{102}+...+\frac{101}{200}-101}{\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}=$

$= \dfrac{(\frac{1}{100}-1)+(\frac{2}{101}-1)+(\frac{3}{102}-1)+...+(\frac{101}{200}-1)}{\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}$

$= \dfrac{\frac{1-100}{100}+\frac{2-101}{101}+\frac{3-102}{102}+...+\frac{101-200}{200}}{\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}$

$=\dfrac{\frac{-99}{100}+\frac{-99}{101}+\frac{-99}{102}+...+\frac{-99}{200}}{\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}$

$=\dfrac{-99\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)}{\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}$

$=-99\cdot \dfrac{\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}{\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}$

$=-99\cdot 1$

$=-99$