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a fost răspuns

Fie a=[tex]\frac{1}{1*2}[/tex]+[tex]\frac{1}{2*3}[/tex]+[tex]\frac{1}{3*4}[/tex]+..............+[tex]\frac{1}{99*100}[/tex]. Demonstrati ca [tex](a+0,01)^{2018}[/tex] ∈ ℕ.[tex][/tex]

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Razvanirina

Știm că [tex]\frac{1}{n(n+1)} =\frac{1}{n}-\frac{1}{n+1}[/tex]

Atunci [tex]a=\frac{1}{1*2} +\frac{1}{2*3}+...+\frac{1}{99*100}= 1-\frac{1}{2} +\frac{1}{2}-\frac{1}{3} +...+\frac{1}{99} -\frac{1}{100} =1-\frac{1}{100}=1-0,01\\\\(a+0.01)^{2018}=(1-0.01+0.01)^{2018}= 1^{2018}=1[/tex]∈|N

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