A) \[2020\]
B) \[1\]
C) \[(2020)!\]
D) \[0\]
Correct Answer: B
Solution :
\[\frac{1}{{{\log }_{2}}n}+\frac{1}{{{\log }_{3}}n}+\frac{1}{{{\log }_{4}}n}+....+\frac{1}{\log {{ & }_{2020}}n}\] \[={{\log }_{n}}2+{{\log }_{n}}3+{{\log }_{n}}4+...+{{\log }_{n}}2020\] \[={{\log }_{n}}(2\times 3\times 4\times .....\times 2020)!\] \[(\because \,n=2020!\,given)\] \[=1\]You need to login to perform this action.
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