Question

$\frac{1}{{\log }_{2}n}+\frac{1}{{\log }_{3}n}+\frac{1}{{\log }_{4}n}+.............+\frac{1}{{\log }_{43}n}$ is equal to

• A. $1$
• B. ${\log }_{43}n!$
• C. ${\log }_{n}43!$
• D. none of these

Answer 1

The correct option is C ${\log }_{n}43!$

$\frac{1}{{\log }_{2}n}+\frac{1}{{\log }_{3}n}+\frac{1}{{\log }_{4}n}+.............+\frac{1}{{\log }_{43}n}$

$={\log }_{n}2+{\log }_{n}3+{\log }_{n}4+.............+{\log }_{n}43$

$={\log }_n(2\times 3\times 4\times ..........\times 43)$

So, the answer is (C)