If $x = {\log _{{2^2}}}2 + {\log _{{2^3}}}{2^2} + {\log _{{2^4}}}{2^3} + \; \ldots .\; + {\log _{{2^{n + 1}}}}{2^n},$ then

$x \ge {\left( {\frac{1}{{n + 1}}} \right)^{1/n}}$

$x \ge {\left( {\frac{n}{{n + 1}}} \right)^{1/n}}$

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JEE • MATHEMATICS • MATHEMATICAL INDUCTION

Mathematical induction practice · Hard

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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x = {\log _{{2^2}}}2 + {\log _{{2^3}}}{2^2} + {\log _{{2^4}}}{2^3} + \; \ldots .\; + {\log _{{2^{n + 1}}}}{2^n},

then

View answer and explanation

Correct option: B

x \ge n{\left( {\frac{1}{{n + 1}}} \right)^{1/n}}

Option B,

x \ge n{\left( {\frac{1}{{n + 1}}} \right)^{1/n}}

, is correct because it is derived by applying AM >= GM inequality to the given logarithmic expression.

Revision notes and key takeaways

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