Question

Ayuda,por favor.
Muy agradecida.

Answer 1

SOLUCIÓN

Hola!! :D

Lo haremos mediante sumatorias, "E" podemos expresarla como:

        $E = \dfrac{1}{2\times6}+ \dfrac{1}{4\times9}+ \dfrac{1}{6\times12}+ \dfrac{1}{8\times15}+...\\\\\\E = \dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{72}+\dfrac{1}{120}+...\\\\\\E = \dfrac{1}{6(1)(1+1)} + \dfrac{1}{6(2)(2+1)}+\dfrac{1}{6(3)(3+1)}+\dfrac{1}{6(4)(4+1)}+...\\\\\\Me \: piden \: 20 \: sumandos\:(L\'imite \: Superior)$

       $\sum_{i=1}^{20}\dfrac{1}{6i(i+1)} = \dfrac{1}{6}-\dfrac{1}{6(20+1)} = \boldsymbol{\dfrac{10}{63}}$

Rpta.  Alternativa A