Răspuns:
Explicație pas cu pas:
[tex]S_n=...=\dfrac{1}{3}*(\dfrac{1}{1}-\dfrac{1}{4})+\dfrac{1}{3}*(\dfrac{1}{4}-\dfrac{1}{7})+...+\dfrac{1}{3}*(\dfrac{1}{3n-2}-\dfrac{1}{3n+1})=\dfrac{1}{3}*(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{3n-2}-\dfrac{1}{3n+1})=\dfrac{1}{3} *(\dfrac{1}{1}- \dfrac{1}{3n+1} )=\dfrac{1}{3}*\dfrac{3n+1-1}{3n+1} =\dfrac{n}{3n+1}[/tex]
