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sa se calc suma 1/ [log in baza 2 din 1 +log in baza 2 din 2 +......+log in baza 2 din 2013]+1/[log in baza 3 din 1+log in baza 3 din 2 +....+log in baza 3 din 2013]+..........+1/[log in baza 2013c din 1 +log in bqza 2013 din 2+....log in baza 2013 din 2013]

Răspuns :

Am atasat o rezolvare.

Vezi imaginea LUCASELA

 

[tex]\displaystyle\\ \frac{1}{\log_21+\log_22+...+\log_2{2013}}+\\\\+\frac{1}{\log_31+\log_32+...+\log_3{2013}}+.....+\\\\ +\frac{1}{\log_{2013}1+\log_{2013}2+...+\log_{2013}{2013}}=\\\\\\=\frac{1}{\log_2{(1\cdot2\cdot...\cdot2013)}}+\frac{1}{\log_3{(1\cdot2\cdot...\cdot2013)}}+\\\\+.....+\frac{1}{\log_{2013}{(1\cdot2\cdot...\cdot2013)}}=\\\\\\=\frac{1}{\log_2{(2013!)}}+\frac{1}{\log_3{(2013!)}}+.....+\frac{1}{\log_{2013}{(2013!)}}=[/tex]

[tex]\displaystyle\\=\log_{(2013!)}2+\log_{(2013!)}3+.....+\log_{(2013!)}(2013)=\\\\=\log_{(2013!)}{(2\cdot3\cdot.....\cdot2013)}=\\\\=\log_{(2013!)}{(2013!)}=\boxed{\bf1}[/tex]