In \[n=(1999)\,!\] then \[\sum\limits_{x=1}^{1999}{{{\log }_{n}}x}\] is equal to [AMU 2002]

In \[n=(1999)\,!\] then \[\sum\limits_{x=1}^{1999}{{{\log }_{n}}x}\] i

1.	In \[n=(1999)\,!\] then \[\sum\limits_{x=1}^{1999}{{{\log }_{n}}x}\] is equal to [AMU 2002]
A.	1
B.	0
C.	\[\sqrt[1999]{1999}\]
D.	-1
Answer» B. 0

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In \[n=(1999)\,!\] then \[\sum\limits_{x=1}^{1999}{{{\log }_{n}}x}\] is equal to [AMU 2002]

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