For every positive integral value of n,{3^n} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To determine the values of $n$ for which ${3^n} > {n^3}$,we test small positive integers:For $n = 1$: ${3^1} = 3$ and ${1^3} = 1$. Since $3 > 1$,t

Vedclass · Accepted Answer

$n \geq 4$

Vedclass · Answer

(C) To determine the values of $n$ for which ${3^n} > {n^3}$,we test small positive integers:For $n = 1$: ${3^1} = 3$ and ${1^3} = 1$. Since $3 > 1$,the condition holds.For $n = 2$: ${3^2} = 9$ and ${2^3} = 8$. Since $9 > 8$,the condition holds.For $n = 3$: ${3^3} = 27$ and ${3^3} = 27$. Here $27 = 27$,so the condition ${3^n} > {n^3}$ does not hold.For $n = 4$: ${3^4} = 81$ and ${4^3} = 64$. Since $81 > 64$,the condition holds.For $n = 5$: ${3^5} = 243$ and ${5^3} = 125$. Since $243 > 125$,the condition holds.By mathematical induction,it can be shown that for all $n \geq 4$,the inequality ${3^n} > {n^3}$ is true.

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

Similar Questions

Prove that $(1 + x)^n \ge (1 + nx)$ for all natural numbers $n,$ where $x > -1.$

Prove by using the principle of mathematical induction that for all $n \in N$,$x^{2n}-y^{2n}$ is divisible by $x+y$.

Using mathematical induction,the numbers $a_n$ are defined by $a_0=1$ and $a_{n+1}=3n^2+n+a_n$ for $n \geq 0$. Then $a_n$ is equal to:

When $P$ is a natural number,then ${P^{n + 1}} + {(P + 1)^{2n - 1}}$ is divisible by

Prove the following by using the principle of mathematical induction for all $n \in N$:
$1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \ldots + (2n - 1)(2n + 1) = \frac{n(4n^2 + 6n - 1)}{3}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module