Question

If $n=100!$, then what is the value of the following?$\frac{1}{{\log }_{2}n}+\frac{1}{{\log }_{3}n}+\frac{1}{{\log }_{4}n}+\dots \dots +\frac{1}{{\log }_{100}n}$

• (A) $0$
• (B) $1$
• (C) $2$
• (D) $3$

Answer 1

The Correct Option is B

Explanation:
Given:$\frac{1}{{\log }_{2}n}+\frac{1}{{\log }_{3}n}+\frac{1}{{\log }_{4}n}+\dots ..+\frac{1}{{\log }_{100}n}$$={\log }_{n}2+{\log }_{n}3+{\log }_{n}4+\dots ..+{\log }_{n}100$$={\log }_n[2\times 3\times 4\times \dots \times 100]$$={\log }_{n}100!$$={\log }_{100!}100!$$=1$Hence, the correct option is (B).