Pogressão Arítimética-Geométrica

evandronunes escreveu:Use a dica:

\frac{1}{3.7}= \frac{1}{4.3}-\frac{1}{4.7}

\frac{1}{5.9}= \frac{1}{4.5}-\frac{1}{4.9}

\frac{1}{7.11}= \frac{1}{4.7}-\frac{1}{4.11}
                    .
                    .
                    .
\frac{1}{(2n+1).(2n+5)}= \frac{1}{4.(2n+1)}-\frac{1}{4.(2n+5)}

Podemos escrever:

\frac{1}{3.7}+\frac{1}{5.9}+\frac{1}{7.11}+ . . . + \frac{1}{(2n+1).(2n+5)}=


\left ( \frac{1}{4.3}-\frac{1}{4.7} \right ) +\left ( \frac{1}{4.5}-\frac{1}{4.9} \right ) + \left ( \frac{1}{4.7}-\frac{1}{4.11} \right ) + . . . + \left ( \frac{1}{4.(2n+1)}-\frac{1}{4.(2n+5)} \right ) =


\frac{1}{4}. \left [ \left ( \frac{1}{3}+\frac{1}{5} +\frac{1}{7}+...+\frac{1}{2n+1} \right ) - \left ( \frac{1}{7} +\frac{1}{9}+\frac{1}{11}+...+\frac{1}{2n+1}+\frac{1}{2n+3} +\frac{1}{2n+5} \right )  \right ] =


\frac{1}{4}.  \left ( \frac{1}{3}+\frac{1}{5}  - \frac{1}{2n+3} -\frac{1}{2n+5} \right )  =


\frac{1}{4}.  \left ( \frac{8}{15} - \frac{4.(n+2)}{(2n+3)(2n+5)} \right )  =


  \frac{2}{15} - \frac{(n+2)}{(2n+3)(2n+5)}=


  \frac{8n^2+17n}{15.(2n+3)(2n+5)}