For positive integer n, $3^{n} < n!$ when

n \geq 6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n > 7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n \geq 7

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

n \leq 7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $n \geq 7$
For positive integer $n$ ,
$3^{n} < n!$
Let $n = 1$
$3^{1} < 1!$
$= 3 < 1$ which is not possible.
Let $n = 2$
$3^{2} < 2!$
$= 9 < 2$ which is not possible.
Let $n = 6$
$3^{6} < 6!$
$= 729 < 720$ which is not possible.
Let $n = 7$
$3^{7} < 7!$
$= 2187 < 5040$ which is possible.
$∴$ For positive integer n, $3^{n} < n!$ when $n \geq 7$

Suggest Corrections

Similar questions

n, \frac{n^{7}}{7} + \frac{n^{5}}{5} + \frac{2 n^{3}}{3} - \frac{n}{105}

n

(n - 1) C_{5} + (n - 1) C_{6} = n C_{7}

n - 7

n

D = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (n - 1)! & (n + 1)! & (n + 3)! / n (n + 1) (n + 1)! & (n + 3)! & (n + 5)! / (n + 2) (n + 3) (n + 3)! & (n + 5)! & (n + 7)! / (n + 4) (n + 5)! \end{matrix} ∣ ∣ ∣ ∣ ∣

\frac{D}{(n - 1)! (n + 1)! (n + 3)!}

n

7^{n} - 3^{n}

4

Prove that $\frac{n^{7}}{7} + \frac{n^{5}}{5} + \frac{n^{3}}{3} + \frac{n^{2}}{2} - \frac{37}{210}$ n is a positive integer for all $n ϵ N$ .

Join BYJU'S Learning Program