32 views

The value of $\dfrac{1}{\log_2 n}+ \dfrac{1}{\log_3 n}+\dfrac{1}{\log_4 n}+ \dots + \dfrac{1}{\log_{2017} n}\:\:($ where $n=2017!)$ is

1. $1$
2. $2$
3. $2017$
4. none of these

recategorized | 32 views

$\frac{1}{log_kn} = {log_nk}$ and $log_na + log_nb + log_nc= log_nabc$. Given n = 2017!. After simplification it would converted to $log_{2017!}2017! = 1$
by (231 points)

+1 vote