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1 vote
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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
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2 Answers

4 votes
$\frac{1}{log_kn} = {log_nk}$ and $log_na + log_nb + log_nc= log_nabc$. Given n = 2017!. After simplification it would converted to $log_{2017!}2017! = 1$
0 votes
Yes , the option A is right because Log of A base B = 1/ Log B base A.

So we can convert the given expression as SUM of( Log 2 base n, Log 3 base n ,----------Log 2017 base n ).

Further it can be converted as product  of Log(2.3.4.5......2017) base n.=Log ( fact 2017) base n. Now n= fact 2017

Hence the given expression evaluate to 1.

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