Question

Demonstrati ca 1/n(n+1) =1/(n+1) oricare ar fi n numar diferit de 0

Answer 1

$\dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1} \\ \\ Pornim din membrul stang si ajungem in membrul drept:\\ \\ Descompunem fractia in fractii simple: \\ \\ Consideram relatia: \dfrac{1}{n(n+1)} = \dfrac{A}{n}+\dfrac{B}{n+1} \quad (*)\\ \\ \\ \dfrac{1}{n(n+1)} = ^{^{\big{n+1)}}}\dfrac{A}{n}+^{^{\big{n)}}}\dfrac{B}{n+1} = \dfrac{A(n+1)+Bn}{n(n+1)} = \\ \\ = \dfrac{An+A+Bn}{n(n+1)} = \dfrac{n(A+B)+A}{n(n+1)} \\ \\ \Rightarrow \dfrac{1}{n(n+1)}= \dfrac{n(A+B)+A}{n(n+1)}$

$Facem sistem:\\ \\ \left\{ \begin{array}{c} A+B = 0 \\ A = 1 \end{array} \right \Rightarrow \left\{ \begin{array}{c} 1+B = 0 \\ A = 1 \end{array} \right \Rightarrow \left\{ \begin{array}{c} B = -1 \\ A = 1 \end{array} \right| \\ \\ \\ Inlocuim in relatia (*): \\ \\ \dfrac{1}{n(n+1)} = \dfrac{A}{n}+\dfrac{B}{n+1} \Rightarrow \dfrac{1}{n(n+1)} = \dfrac{1}{n}+\dfrac{-1}{n+1} \Rightarrow\\ \\ \Rightarrow \boxed{\dfrac{1}{n(n+1)} = \dfrac{1}{n}-\dfrac{1}{n+1}}$