For natural number n, $2^{n} (n - 1)! < n^{n}$ , if.

n < 2

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n > 2

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n \geq 2

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Never

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Solution

The correct option is D $n > 2$
$2^{n} (n - 1)! < n^{n}$
If $n = 1$
Then, $2^{1} (1 - 1)! < 1$
$\Rightarrow 2 \times 0! < 1$
$\Rightarrow 2 \times 1 < 1$ which is not possible
If $n = 2$
Then, $2^{2} (2 - 1)! < 2^{2}$
$\Rightarrow 4 \times 1! < 4$ which is not possible.
If $n = 3$ then,
$2^{n} (n - 1)! < n^{n}$
$2^{3} (3 - 1)! < 3^{3}$
$\Rightarrow 4 \times 2! < 27$
$16 < 27$ which is possible.
Therefore, for natural number n, $2^{n} (n - 1)! < n^{n}$ if $n > 2$

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Similar questions

For natural number n, $2^{n}$ (n-1) ! < $n^{n}$ , if

n \geq 2, \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \dots \dots + \frac{1}{\sqrt{n}} > \sqrt{n}

n \geq 2, \sqrt{n (n + 1)} < n + 1

Statement I

n \geq 2

\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}} > \sqrt{n}

Statement II

n \geq 2

\sqrt{n (n + 1)} < n + 1

n \geq 2,

\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}} > \sqrt{n}

\sqrt{n (n + 1)} < n + 1

For natural number n, $(n!)^{2}$ > $n^{n}$ , if

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