Question

If $n>1$, then value of $\frac{1}{{\log }_{2}n}+\frac{1}{{\log }_{3}n}+\frac{1}{{\log }_{4}n}+......+\frac{1}{{\log }_{50}n}$ is

• A. $\frac{1}{{\log }_{n}50!}$
• B. $\frac{1}{{\log }_{n!}50}$
• C. $1$
• D. $\frac{1}{{\log }_{50!}n}$

Answer 1

The correct option is D $\frac{1}{{\log }_{50!}n}$

We have,

$\frac{1}{{\log }_{2}n}+\frac{1}{{\log }_{3}n}+\frac{1}{{\log }_{4}n}+......+\frac{1}{{\log }_{50}n}{\log }_{n}2+{\log }_{n}3+{\log }_{n}4+.......+{\log }_{n}50=\log 50!=\frac{1}{{\log }_{50!}n}$

Hence, this is the answer.