A) 1
B) 0
C) \[\sqrt[1999]{1999}\]
D) - 1
Correct Answer: A
Solution :
\[\sum\limits_{x=1}^{1999}{{{\log }_{n}}x}\] \[={{\log }_{(1999)!}}1+{{\log }_{(1999)!}}2+.......+{{\log }_{(1999)!}}1999\] \[={{\log }_{(1999)!}}(1.2.3.......1999)={{\log }_{(1999)!}}(1999)!=1\].
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