If P(n): 3n < n!, n N, Then P(n) is true for: - Vidyadip

MCQ

If $P(n): 3n < n!, n\ \epsilon\ N,$ Then $P(n)$ is true for:

✓
$n ≥ 7$
B
$n ≥ 3$
C
$n ≥ 6$
D
all

✓

Answer

Correct option: A.

$n ≥ 7$

Concept:
$\text{n}!=\text{n}\times(\text{n}-1)\times(\text{n}-2).....\times3\times2\times1$
Given:
$P(n): 3n < n!$
This can be solved directly by hit and trial method, putting the option in expression and checking its validity
We will choose first the smallest number from the options
Putting $n = 3$
$ P(n) = 3n < n! = 3^3 < 3! $
$\Rightarrow 27 ≮ 3 \times 2 \times 1, 27 ≮ 6 ,$
Hence Option $2$ is wrong
Putting $n = 6$
$P(n) = 3n < n! = 63 < 6!$
$P(n) = 3n < n! = 3^6 < 6! , 1029 < 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$1029 ≮ 720$ Hence Option $3$ is wrong
Put $n = 7$
$P(n) = 3n < n! = 3^7 < 7! , 2187 < 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$2187 < 5040,$ Option $1$ satisfies the given expression,
Hence the correct answer is option $1.$

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