Question

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Answer 1

Răspuns:

Explicație pas cu pas:

Calculăm suma...

$a) ~\dfrac{1}{1*2}+ \dfrac{1}{2*3}+...+\dfrac{1}{99*100}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{1}-\dfrac{1}{100}=\dfrac{100}{1100}-\dfrac{1}{100}=\dfrac{99}{100}.~~Am~obtinut~~\\0,5<\dfrac{99}{100}<1,~~=>~\dfrac{5}{10}<\dfrac{99}{100}<\dfrac{100}{100} ,~=>~\dfrac{50}{100}<\dfrac{99}{100}<\dfrac{100}{100} ,~adevarat.$

Deci, este adevărată și relația dată.

$b)~~\dfrac{2}{1*3}+\dfrac{2}{3*5}+\dfrac{2}{5*7}+...+\dfrac{2}{99*101}=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}=\dfrac{1}{1}-\dfrac{1}{101} =\dfrac{100}{101}.~~Deci~obtinem:\\\dfrac{^{101)}2}{5}<\dfrac{^{5)}100}{101}<1,~~=>~~\dfrac{202}{505} <\dfrac{500}{505} < \dfrac{505}{505} ,~~adevarat.$