Question

un exercitiu mai deosebit...
Raspunsul este 1

Answer 1

 

$\displaystyle\\ \text{Folosim formulele:}\\\\ \log_ax+\log_ay=\log_axy~~~\forall~x,y>0\\\\1\cdot2\cdot3\cdot4\cdot...\cdot n=2\cdot3\cdot4\cdot...\cdot n=n!~~\text{(n factorial)}\\\\\log_ab=\frac{1}{\log_ba}~~~\forall~a,b>0;~a,b\neq1\\\\\\\text{Rezolvare:}\\\\S=\frac{1}{\log_21+\log_22+...+\log_22018}+\\\\ +\frac{1}{\log_31+\log_32+...+\log_32018}+...\\\\...+\frac{1}{\log_{2018}1+\log_{2018}2+...+\log_{2018}2018}$

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$\displaystyle\\S=\frac{1}{\log_2(1\cdot2\cdot ... \cdot2018)}+\frac{1}{\log_3(1\cdot2\cdot ... \cdot2018)}+...+\frac{1}{\log_{2018}(1\cdot2\cdot ... \cdot2018)}\\\\S=\frac{1}{\log_2(2018!)}+\frac{1}{\log_3(2018!)}+...+\frac{1}{\log_{2018}(2018!)}\\\\S=\log_{2018!}(2)+\log_{2018!}(3)+...+\log_{2018!}(2018)\\\\S=\log_{2018!}(2\cdot3\cdot...2018)\\\\S=\log_{2018!}(2018!)\\\\\boxed{\bf~S=1}$

 

 

Answer 2

Răspuns:

Explicație pas cu pas:

Rezolvarea este in imaginea de mai jos. Mult succes!