A) \[1\]
B) \[2\]
C) \[3\]
D) \[4\]
Correct Answer: A
Solution :
\[\frac{1}{{{\log }_{2}}n!}+\frac{1}{{{\log }_{3}}n!}+\frac{1}{{{\log }_{4}}n!}+....+\frac{1}{{{\log }_{2002}}n!}\] \[=\frac{\log \,2}{\log n!}+\frac{\log \,3}{\log \,n!}+.....+\frac{\log 2002}{\log n!}\] \[=\frac{\log (2.3.4....2002)}{\log n!}=\frac{\log 2002!}{\log n!}\] \[=\frac{\log \,2002!}{\log \,2002!}\] \[(\because \,\,\,n=2002\,given)\] \[=1\]
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