KCET · Maths · Sequences and Series
If \(P(n): 2^{n} < n\) ! Then the smallest positive integer for which \(P(n)\) is true if
- A 2
- B 3
- C 4
- D 5
Answer & Solution
Correct Answer
(C) 4
Step-by-step Solution
Detailed explanation
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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