KCET · Maths · Mathematical Induction
If \( P(n): 2^{n} < n ! \) then the smallest positive integer for which \( P(n) \) is true, is
- A \( 03 \)
- B \( 05 \)
- C \( 02 \)
- D \( 04 \)
Answer & Solution
Correct Answer
(D) \( 04 \)
Step-by-step Solution
Detailed explanation
\(\\) P(\mathrm{n}): 2^{\wedge}\{\mathrm{n}\}\( Since
\)P(1): 2 < 1\( is false.
\)P(2): 2^{2} < 2 !\( is false
\)P(3): 2^{3} < 3 !\( is false
\)P(4): 2^{4} < 4 !\( is true.
\)P(5): 2^{5} < 5 !\( is true.
So, the smallest positive integer is \)4 .$
\)P(1): 2 < 1\( is false.
\)P(2): 2^{2} < 2 !\( is false
\)P(3): 2^{3} < 3 !\( is false
\)P(4): 2^{4} < 4 !\( is true.
\)P(5): 2^{5} < 5 !\( is true.
So, the smallest positive integer is \)4 .$
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