Question

(UFRGS) A soma ㏒2/4 + ㏒4/5+........+ ㏒19/20 é igual a:

Answer 1

Olá!

Pensei no seguinte, veja:

$\\ \mathsf{\log \left ( \frac{2}{4} \right ) + \log \left ( \frac{4}{5} \right ) + ... + \log \left ( \frac{19}{20} \right ) =} \\\\\\ \mathsf{\log \left ( \frac{1}{2} \right ) + \log \left ( \frac{4}{5} \right ) + \log \left ( \frac{7}{8} \right ) + \log \left ( \frac{10}{11} \right ) + \log \left ( \frac{13}{14} \right ) + \log \left ( \frac{16}{17} \right ) + \log \left ( \frac{19}{20} \right ) =} \\\\\\ \mathsf{(\log 1 - \log 2) + (\log 4 - \log 5) + (\log 7 - \log 8) + (\log 10 - \log 11) + ... + (\log 19 - \log 20) =} \\\\ \mathsf{(\log 1 + \log 4 + \log 7 + \log 10 + ... + \log 19) - (\log 2 + \log 5 + \log 8 + \log 11 + ... + \log 20) =} \\\\ \mathsf{\log (1 \cdot 4 \cdot 7 \cdot 10 \cdot ... \cdot 19) - \log (2 \cdot 5 \cdot 8 \cdot 11 \cdot ... \cdot 20) =}$

$\\ \mathsf{\log (1 \cdot 4 \cdot 7 \cdot 10 \cdot ... \cdot 19) - \log (2 \cdot 5 \cdot 8 \cdot 11 \cdot ... \cdot 20) =} \\\\ \mathsf{\log \left ( \frac{1 \cdot 4 \cdot 7 \cdot 10 \cdot ... \cdot 19}{2 \cdot 5 \cdot 8 \cdot 11 \cdot ... \cdot 20} \right ) =} \\\\\\ \mathsf{\log \left ( \frac{1 \cdot 4 \cdot 7 \cdot 10 \cdot ... \cdot 19}{2 \cdot 5 \cdot 8 \cdot 11 \cdot ... \cdot 20} \right ) =} \\\\\\ \mathsf{\log \left ( \frac{13 \cdot 19}{5 \cdot 11 \cdot 17} \right ) =} \\\\ \boxed{\mathsf{\log 247 - \log 935}}$

Espero ter ajudado!

Boa questão!!