Matematică
silviacherbeleata
2015-11-12 08:02:54
Demonstrati ca 1/n(n+1) =1/(n+1) oricare ar fi n numar diferit de 0
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Marilusa
2015-11-12 14:18:50

[latex]dfrac{1}{n(n+1)} = dfrac{1}{n} - dfrac{1}{n+1} \ \ $Pornim din membrul stang si ajungem in membrul drept:\ \ Descompunem fractia in fractii simple: \ $\ $Consideram relatia: dfrac{1}{n(n+1)} = dfrac{A}{n}+dfrac{B}{n+1} quad (*)\ \ \ dfrac{1}{n(n+1)} = ^{^{ig{n+1)}}}dfrac{A}{n}+^{^{ig{n)}}}dfrac{B}{n+1} = dfrac{A(n+1)+Bn}{n(n+1)} = \ \ = dfrac{An+A+Bn}{n(n+1)} = dfrac{n(A+B)+A}{n(n+1)} \ \ Rightarrow dfrac{1}{n(n+1)}= dfrac{n(A+B)+A}{n(n+1)}[/latex] [latex] $ Facem sistem:\ \ left{ egin{array}{c} A+B = 0 \ A = 1 end{array} ight Rightarrow left{ egin{array}{c} 1+B = 0 \ A = 1 end{array} ight Rightarrow left{ egin{array}{c} B = -1 \ A = 1 end{array} ight| \ \ \ $Inlocuim in relatia $ (*): \ \ dfrac{1}{n(n+1)} = dfrac{A}{n}+dfrac{B}{n+1} Rightarrow dfrac{1}{n(n+1)} = dfrac{1}{n}+dfrac{-1}{n+1} Rightarrow\ \ Rightarrow oxed{dfrac{1}{n(n+1)} = dfrac{1}{n}-dfrac{1}{n+1}} [/latex]

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