The smallest positive integer for which the statement $3^{n + 1} < 4^{n}$ holds is

1

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

2

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

3

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

4

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

Open in App

Solution

The correct option is C $4$
Let $P (n) : 3^{n + 1} < 4^{n}$

$P (1) : 3^{2} < 4$ which is false

$P (2) : 3^{3} < 4^{2}$ which is false

$P (3) : 3^{4} < 4^{3}$ which is false

$P (4) : 3^{5} < 4^{4}$ which is true

Suggest Corrections

Similar questions

n

n! < {(\frac{n + 1}{2})}^{n}

2^{n} - (n - 1) 2^{n - 2} + \frac{(n - 2) (n - 3)}{⌊ 2} - \frac{(n - 3) (n - 4) (n - 5)}{⌊ 3} 2^{n - 6} + . . . . .;

3

1 - (n - 1) + \frac{(n - 2) (n - 2)}{⌊ 2} - \frac{(n - 3) (n - 4) (n - 5)}{⌊ 3} + . . . . = (- 1)^{n}

1 + \frac{1}{2} + \frac{1}{3} + . . . . + \frac{1}{n} \geq 4

b = a + 1

b^{n} - (n - 1) a b^{n - 2} + \frac{(n - 2) (n - 3)}{⌊ 2} a^{2} b^{n - 4} - \frac{(n - 3) (n - 4) (n - 5)}{⌊ 3} a^{3} b^{n - 6} + . . . .

n

{(\frac{1 + i \sqrt{3}}{1 - i \sqrt{3}})}^{n} = 1

Join BYJU'S Learning Program