Explicație pas cu pas:
[tex]\sum\limits_{k=1}^n \dfrac{1}{k(k+1)(k+2)} = \sum\limits_{k=1}^n \dfrac{(k+2)-k}{k(k+1)(k+2)}\cdot \dfrac{1}{2} = \\ \\ = \dfrac{1}{2}\sum\limits_{k=1}^n \left[\dfrac{k+2}{k(k+1)(k+2)}-\dfrac{k}{k(k+1)(k+2)}\right]} = \\ \\= \dfrac{1}{2}\sum\limits_{k=1}^n\left[ \dfrac{1}{k(k+1)} - \dfrac{1}{(k+1)(k+2)}\right] = \\ \\= \dfrac{1}{2}\sum\limits_{k=1}^n\dfrac{1}{k(k+1)}-\dfrac{1}{2}\sum\limits_{k=1}^n\dfrac{1}{(k+1)(k+2)}=[/tex]
[tex]= \dfrac{1}{2}\cdot\Big[ \dfrac{1}{1\cdot2}+\sum\limits_{k=1}^{n-1}\dfrac{1}{(k+1)(k+2)}\Big] -\dfrac{1}{2} \cdot\Big[ \sum\limits_{k=1}^{n-1}\dfrac{1}{(k+1)(k+2)}+\dfrac{1}{(n+1)(n+2)}\Big]=\\ \\ =\dfrac{1}{2}\cdot \Big[ \dfrac{1}{1\cdot 2}-\dfrac{1}{(n+1)(n+2)}\Big][/tex]
