A) 0
B) 1
C) 10
D) \[{{10}^{3}}\]
Correct Answer: B
Solution :
Given that \[n=1000!\] Now, \[\frac{1}{{{\log }_{2}}n}+\frac{1}{{{\log }_{3}}n}+...+\frac{1}{{{\log }_{1000}}n}\] \[={{\log }_{n}}2+{{\log }_{n}}3+....+{{\log }_{n}}1000\] \[={{\log }_{n}}2.3.4....1000\] \[={{\log }_{n}}(1000!)=lo{{g}_{n}}n=1\]You need to login to perform this action.
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