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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
in Numerical Ability by Boss (17.5k points)
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$\frac{1}{log_kn} = {log_nk}$ and $log_na + log_nb + log_nc= log_nabc$. Given n = 2017!. After simplification it would converted to $log_{2017!}2017! = 1$
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