[tex]\text{Operatia } "*" \text{care satisface acele proprietati este operatia de impartire.} \\ \\ \sum\limits_{k=1} ^{2017} \Big[\dfrac{1}{k}* (k+1)\Big] = \sum\limits_{k=1} ^{2017} \Big[ \dfrac{1}{k}: (k+1)\Big] = \sum\limits_{k=1} ^{2017} \dfrac{\dfrac{1}{k}}{k+1}= \\ \\ = \sum\limits_{k=1} ^{2017} \dfrac{1}{k(k+1)} =\sum\limits_{k=1} ^{2017} \Big[ \dfrac{1}{k} - \dfrac{1}{k+1}\Big] =\sum\limits_{k=1} ^{2017} \dfrac{1}{k} - \sum\limits_{k=1} ^{2017} \dfrac{1}{k+1} =[/tex]
[tex]\Big[\dfrac{1}{1} +\sum\limits_{k=1} ^{2016}\dfrac{1}{k+1}\Big] - \Big[\sum\limits_{k=1} ^{2016}\dfrac{1}{k+1} + \Big(\dfrac{1}{2017+1}\Big)\Big] = \\ \\ = 1 - \dfrac{1}{2018} +\sum\limits_{k=1} ^{2016}\dfrac{1}{k+1} -\sum\limits_{k=1} ^{2016}\dfrac{1}{k+1} = \\ \\ = 1- \dfrac{1}{2018} = \dfrac{2018-1}{2018} = \dfrac{2017}{2018} \rightarrow \text{f) corect}[/tex]
