Răspuns:
Explicație pas cu pas:
[tex]\dfrac{1}{1*3}+ \dfrac{1}{3*5}+...+\dfrac{1}{(2n-1)(2n+1)}=\dfrac{1}{2}*(\dfrac{1}{1}-\dfrac{1}{3})+\dfrac{1}{2}*(\dfrac{1}{3}-\dfrac{1}{5})+...+\dfrac{1}{2}*(\dfrac{1}{2n-1}-\dfrac{1}{2n+1})=\dfrac{1}{2}*(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1})=\dfrac{1}{2}*(\dfrac{1}{1}-\dfrac{1}{2n+1})=\\ = \dfrac{1}{2}*\dfrac{2n+1-1}{2n+1}=\dfrac{n}{2n+1}[/tex]
A fost aplicată identitatea
[tex]\dfrac{1}{n*(n+k)}=\dfrac{1}{k}*( \dfrac{1}{n}-\dfrac{1}{(n+k)})[/tex]
