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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Basics of Mathematics

NDA 2018 · Mathematics · Basics of Mathematics

If n=(2017)! , then what is 1 _2 n + 1 _3 n + 1 _4 n + .+ 1 _ 2017...

If n=(2017)!n=(2017)!, then what is 1log⁡2n+1log⁡3n+1log⁡4n+….+1log⁡2017n\frac{1}{\log _2 \mathrm{n}}+\frac{1}{\log _3 \mathrm{n}}+\frac{1}{\log _4 \mathrm{n}}+\ldots .+\frac{1}{\log _{2017} \mathrm{n}} equal to

Options

  1. A0
  2. B1
  3. Cn2\frac{\mathrm{n}}{2}
  4. Dn

Correct answer

B. 1

Step-by-step solution

n=(2017)!1log⁡2n+1log⁡3n+1log⁡4n+….+1log⁡2017n=1log⁡nlog⁡2+1log⁡nlog⁡3+1log⁡nlog⁡4+….+1log⁡nlog⁡2017(∵log⁡ab=log⁡blog⁡a)=log⁡2log⁡n+log⁡3log⁡n+log⁡4log⁡n+….+log⁡2017log⁡n=log⁡2+log⁡3+log⁡4+….+log⁡2017log⁡n=log⁡(2⋅3⋅4…2017)log⁡n(∵log⁡a+log⁡b+log⁡c+…log⁡( a.b.c..... )=log⁡(2017!)log⁡n=log⁡nlog⁡n=1\begin{aligned} & \mathrm{n}=(2017)! \\ & \frac{1}{\log _2 n}+\frac{1}{\log _3 n}+\frac{1}{\log _4 n}+\ldots .+\frac{1}{\log _{2017} n} \\ & =\frac{1}{\frac{\log n}{\log 2}}+\frac{1}{\frac{\log n}{\log 3}}+\frac{1}{\frac{\log n}{\log 4}}+\ldots .+\frac{1}{\frac{\log n}{\log 2017}} \\ & \left(\because \log _{\mathrm{a}} \mathrm{ b}=\frac{\log _{\mathrm{b}}}{\log _{\mathrm{a}}}\right) \\ & =\frac{\log 2}{\log n}+\frac{\log 3}{\log n}+\frac{\log 4}{\log n}+\ldots .+\frac{\log 2017}{\log n} \\ & =\frac{\log 2+\log 3+\log 4+\ldots .+\log 2017}{\log n} \\ & =\frac{\log (2 \cdot 3 \cdot 4 \ldots 2017)}{\log _{\mathrm{n}}} \quad(\because \log a+\log b+\log c+\ldots \log (\text{ a.b.c..... }) \\ & =\frac{\log (2017!)}{\log _{\mathrm{n}}}=\frac{\log _{\mathrm{n}}}{\log _{\mathrm{n}}}=1 \end{aligned}
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