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Question:
Answered & Verified by Expert
If $\mathrm{n}=(2020)$ ! then
$$
\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\frac{1}{\log _{4} n}+\ldots+\frac{1}{\log _{2020} n}
$$
is equal to
Options:
$$
\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\frac{1}{\log _{4} n}+\ldots+\frac{1}{\log _{2020} n}
$$
is equal to
Solution:
2744 Upvotes
Verified Answer
The correct answer is:
1
$\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\frac{1}{\log _{4} n}+\ldots+\frac{1}{\log _{2020} n}$
$=\log _{n} 2+\log _{n} 3+\log _{n} 4+\ldots+\log _{n} 2020$
$=\log _{n}(2 \times 3 \times 4 \times \ldots \times 2020)$
$=\log _{(2020) !}(2020) ! \quad(\because n=2020 !$ given $)$
$=1$
$=\log _{n} 2+\log _{n} 3+\log _{n} 4+\ldots+\log _{n} 2020$
$=\log _{n}(2 \times 3 \times 4 \times \ldots \times 2020)$
$=\log _{(2020) !}(2020) ! \quad(\because n=2020 !$ given $)$
$=1$
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