Question

If If $n =(2020)$, then $\frac {1}{\log_2n}+\frac {1}{\log_3n}+\frac {1}{\log_4n}+............+\frac {1}{\log_{2020} n}$

• A. 0
• B. 2020
• C. 1
• D. $\angle {2020}$

Answer 1

Answer: (C)

1

Answer 2

$\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\frac{1}{\log _{4} n}+\ldots+\frac{1}{\log _{2020} n}$ $=\log _{n} 2+\log _{n} 3+\log _{n} 4+\ldots+\log _{n} 2020$ $=\log _{n}(2 \times 3 \times 4 \times \ldots \times 2020)$ $=\log _{(2020) 1}(2020) ! \,\,\,\,\, (\because n=2020 !$ given $)$ $=1$