For every positive integral value of $n$,${3^n} > {n^3}$ when

  • A
    $n > 2$
  • B
    $n \geq 3$
  • C
    $n \geq 4$
  • D
    $n < 4$

Explore More

Similar Questions

For all $n \ge 1,$ prove that $1^{2}+2^{2}+3^{2}+4^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}$

Use the Principle of Mathematical Induction to prove that $\cos \theta \cos 2 \theta \cos 2^{2} \theta \ldots \cos 2^{n-1} \theta = \frac{\sin 2^{n} \theta}{2^{n} \sin \theta}$ for all $n \in N$.

Difficult
View Solution

Prove the following by using the principle of mathematical induction for all $n \in N$:
$\frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \frac{1}{7 \times 9} + \ldots + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}$

Let $P(n): 2+2^2+2^3+\ldots+2^n=2^{n+1}-2, n \in N$. Then,

Prove the statement by the Principle of Mathematical Induction: $n(n^{2}+5)$ is divisible by $6$,for each natural number $n$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo