calculati [Sn] , unde Sn= 1/13+1/35+...+1/(2n-1)(2n+1)

[tex] S_n = \sum\limits_{k=1}^n \dfrac{1}{(2k-1)(2k+1)} = \\ \\ =\sum\limits_{k=1}^n \dfrac{1}{2}\cdot \dfrac{2}{(2k-1)(2k+1)} = \\ \\ = \sum\limits_{k=1}^n \dfrac{1}{2}\cdot \dfrac{(2k+1)-(2k-1)}{(2k-1)(2k+1)}= \\ \\ = \dfrac{1}{2} \cdot \sum\limits_{k=1}^n \left(\dfrac{2k+1}{(2k-1)(2k+1)} - \dfrac{2k-1}{(2k-1)(2k+1)} \right) = \\ \\ = \dfrac{1}{2} \cdot \sum\limits_{k=1}^n \left(\dfrac{1}{2k-1} - \dfrac{1}{2k+1} \right) =\\ \\ = \dfrac{1}{2} \cdot \left( \sum\limits_{k=1}^n \dfrac{1}{2k-1} - \sum\limits_{k=1}^n \dfrac{1}{2k+1}\right) [/tex]

[tex] = \dfrac{1}{2} \cdot \left( 1 + \dfrac{1}{3} + \dfrac{1}{5} + ... + \dfrac{1}{2n-1} - \dfrac{1}{3} - \dfrac{1}{5} -$ $ \\ \\ ...\left - \dfrac{1}{2n-1} - \dfrac{1}{2n+1}\right) = \\ \\ = \dfrac{1}{2} \cdot \left(1 - \dfrac{1}{2n+1} \right) =\dfrac{1}{2} \cdot \dfrac{2n+1-1}{2n+1} = \dfrac{n}{2n+1} [/tex]