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If $P(n): 2^{n} < n$ ! Then the smallest positive integer for which $P(n)$ is true if
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Verified Answer
The correct answer is:
4
We have,
$\begin{aligned}
&P(n)=2^{n} < n ! \\
&P(1)=2 < 1 ! \text { False }
\end{aligned}$
$\begin{aligned}
&P(2)=2^{2} < 2 ! \text { False } \\
&P(3)=2^{3} < 3 \text { ! False } \\
&P(4)=2^{4} < 4 \text { ! True }
\end{aligned}$
$\therefore$ The smallest position integer for which $P(n)$ is true for $n=4$.
$\begin{aligned}
&P(n)=2^{n} < n ! \\
&P(1)=2 < 1 ! \text { False }
\end{aligned}$
$\begin{aligned}
&P(2)=2^{2} < 2 ! \text { False } \\
&P(3)=2^{3} < 3 \text { ! False } \\
&P(4)=2^{4} < 4 \text { ! True }
\end{aligned}$
$\therefore$ The smallest position integer for which $P(n)$ is true for $n=4$.
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