A) 2006
B) 2005
C) 2005!
D) 1
E) 0
Correct Answer: D
Solution :
Given that, \[n=2006!\] \[\therefore \] \[\frac{1}{{{\log }_{2}}n}+\frac{1}{{{\log }_{3}}n}+.....+\frac{1}{{{\log }_{2006}}n}\] \[={{\log }_{n}}2+{{\log }_{n}}3+.....+{{\log }_{n}}2006\] \[={{\log }_{n}}(2.3.4......2006)\] \[={{\log }_{n}}(2006!)={{\log }_{n}}n=1\]You need to login to perform this action.
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