Question

If If $n=(2020)$, then $\frac{1}{{\log }_{2}n}+\frac{1}{{\log }_{3}n}+\frac{1}{{\log }_{4}n}+............+\frac{1}{{\log }_{2020}n}$

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

Answer 1

Answer: (C)

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