Sum of Logs

Since $32 = (\frac {1}{2})^{-5} \Rightarrow \log_{0.5} {32} = -5$ . Similarly, $\log_{0.5} {8} = -3$ .

Therefore,

$\sum _{n=1} ^4 {\frac {\log_{0.5} {32} ^{\frac{1}{n} } }{\log_{0.5} {8}^{n+2} }} = \sum _{n=1} ^4 {\frac {-5} {-3n(n+2)}} = \frac {5} {3} \sum _{n=1} ^4 {\frac {1} {n(n+2)}}$

$\quad = \frac {5}{3} \left( \frac {1} {3} + \frac {1} {8} + \frac {1} {15} + \frac {1} {24} \right) = \frac {5} {3} \times \frac {40+15+8+5} {120} = \frac {5} {3} \times \frac {68} {120} = \boxed{\frac {17} {18}}$

$k\quad =\frac { \log _{ { 2 }^{ -1 }\quad }{ { 2 }^{ \frac { 5 }{ n } } } }{ \log _{ { { 2 }^{ -1 }\quad }^{ } }{ { 2 }^{ 3n+6 } } } ,\quad now\quad by\quad definition\quad of\quad log\quad we\quad can\\ easily\quad obtain\quad the\quad numerator\quad =\quad \frac { -5 }{ n } \quad and\\ denominator\quad =\quad -3n-6\\ simplifting\quad we\quad get\quad k\quad =\quad \frac { 5 }{ 3n(n+2) } \\ now\quad get\quad 4\quad values\quad of\quad k\quad for\quad n\quad =\quad 1\quad to\quad 4\quad and\quad add\quad all\quad the\quad \\ values\quad to\quad finally\quad get\quad the\quad answer\quad as\quad \frac { 17 }{ 18 }$