The values of the natural numbers n for which | vedclass

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

Question

(A) We test the inequality $2^n > 2n + 1$ for natural numbers $n = 1, 2, 3, 4, \dots$ For $n = 1$: $2^1 = 2$ and $2(1) + 1 = 3$. Since $2 > 3$ is fals

Vedclass · Accepted Answer

For $n \ge 3$

Vedclass · Answer

(A) We test the inequality $2^n > 2n + 1$ for natural numbers $n = 1, 2, 3, 4, \dots$ For $n = 1$: $2^1 = 2$ and $2(1) + 1 = 3$. Since $2 > 3$ is false,the inequality does not hold for $n = 1$. For $n = 2$: $2^2 = 4$ and $2(2) + 1 = 5$. Since $4 > 5$ is false,the inequality does not hold for $n = 2$. For $n = 3$: $2^3 = 8$ and $2(3) + 1 = 7$. Since $8 > 7$ is true,the inequality holds for $n = 3$. For $n = 4$: $2^4 = 16$ and $2(4) + 1 = 9$. Since $16 > 9$ is true,the inequality holds for $n = 4$. By mathematical induction,it can be shown that the inequality holds for all $n \ge 3$.

The values of the natural numbers $n$ for which the inequality $2^n > 2n + 1$ is valid are:

Similar Questions

Prove the following by using the principle of mathematical induction for all $n \in N$:
$\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \ldots + \frac{1}{(3n-2)(3n+1)} = \frac{n}{3n+1}$

If $n$ is a positive integer,then $2 \cdot 4^{2n+1} + 3^{3n+1}$ is divisible by

Let $P(n)$ denote the statement that $n^2 + n$ is odd. It is seen that $P(n) \Rightarrow P(n + 1)$. $P(n)$ is true for all:

Prove the statement by the Principle of Mathematical Induction: $n^{2} < 2^{n}$ for all natural numbers $n \geq 5$.

If $P(n): 2^{n} < n!$,then the smallest positive integer $n$ for which $P(n)$ is true is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module