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

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

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

JEE MATHEMATICS MCQ: MATHEMATICAL INDUCTION

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Analyzing +4 | -1

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.

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\frac{1^2}{3}+\frac{1^2+2^2}{5}+\frac{1^2+2^2+3^2}{7}+\ldots

n

P\left( n \right):2 + 4 + 6 \ldots + \left( {2n} \right),\;n \in N,

\begin{array}{l} P\left( k \right) = k\left( {k + 1} \right) + 2\mathrm{implies}\\ P\left( k \right) = \left( {k + 1} \right)\left( {k + 2} \right) + 2 \end{array}

k \in N.

P\left( n \right) = n\left( {n + 1} \right) + 2

n

n

a\left( n \right) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{\left( {{2^n}} \right) - 1}}.

n(n+1)(n+2)(n+3)

n \in N

12 {~N}

8 {~m}

6 {~m}