Question

If $n=\left(2017\right)!$, then what is
$\frac{1}{{\log }_{2}n}+\frac{1}{{\log }_{3}n}+\frac{1}{{\log }_{4}n}+....+\frac{1}{{\log }_{2017}n}$ equal to

• A. $0$
• B. $1$
• C. $\frac{n}{2}$
• D. $n$

Answer 1

The correct option is B $1$

If $n=\left(2017\right)!$

$\frac{1}{{\log }_{2}n}+\frac{1}{{\log }_{3}n}+...+\frac{1}{{\log }_{2017}n}.....\left(1\right)$

Now we know that $\frac{1}{{\log }_{a}b}={\log }_{b}a$

$\therefore$ we can rewrite Equation 1 as

$\frac{1}{{\log }_{2}n}+\frac{1}{{\log }_{3}n}+...+\frac{1}{{\log }_{2017}n}$

$={\log }_{n}2+{\log }_{n}3+{\log }_{n}4+...+{\log }_{n}2017$

$={\log }_n(2.3.4.....2017)$ as $[{\log }_{a}b+{\log }_{a}c+{\log }_a(b\times c\left)\right]$

$n=\left(2017\right)!$

$\therefore {\log }_n\left(\mathrm{2.3.4....2017}\right)={\log }_{2017!}\left(2017\right)!=1$

Hence,

$\frac{1}{{\log }_{2}n}+\frac{1}{{\log }_{3}n}+...+\frac{1}{{\log }_{2017}n}=1$