Question

Pessoal, alguém pode me ajudar nessa questão?

Answer 1

$\displaystyle{\sum_{n=1}^4\dfrac{log_{\frac{1}{2}}\sqrt[n]{32}}{log_{\frac{1}{2}}8^{n+2}} =\\\displaystyle{\sum_{n=1}^4\dfrac{log_{2^{-1}}\sqrt[n]{32}}{log_{2^{-1}}8^{n+2}} = \displaystyle{\sum_{n=1}^4\dfrac{log_{2}\sqrt[n]{32}}{log_{2}8^{n+2}} \cdot\dfrac{-1}{-1}$

$\displaystyle{\sum_{n=1}^4\dfrac{log_{2}32^{\frac{1}{n}}}{log_{2}{8}^{n+2}} =\displaystyle{\sum_{n=1}^4\cdot\dfrac{\dfrac{1}{n}\cdot\overbrace{log_{2}32}^5}{(n+2).\underbrace{log_28}_3}=\dfrac{5}{3}\cdot\displaystyle{\sum_{n=1}^4\dfrac{1}{n.(n+2)}}$

$\dfrac{5}{3}\cdot\displaystyle{\sum_{n=1}^4\dfrac{1}{n.(n+2)} = \dfrac{5}{3}.\left(\dfrac{1}{1.(1+2)}+\dfrac{1}{2.(2+2)}+\dfrac{1}{3.(3+2)}+\dfrac{1}{4.(4+2)}\right)$

$= \dfrac{5}{3}.\left(\dfrac{1}{3}+\dfrac{1}{8}+\dfrac{1}{15}+\dfrac{1}{24}\right) = \dfrac{5}{3}.\dfrac{17}{30} = \dfrac{17}{3.6} = \large{\boxed{\dfrac{17}{18}}}$

Resposta: D)