/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Prove that \(2^{n}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 14

Prove that \(2^{n}

Short Answer

Expert verified
The inequality \(2^{n}<n !\) for all \(n \geq 4, n \in \mathbb{N}\) is proven to be true by using mathematical induction. The base case \(n=4\) is true and the inductive step from \(n=k\) to \(n=k+1\) is proven to be true. Hence, the inequality holds for all natural numbers \(n \geq 4\).

Step by step solution

01

Base Case

First, validate the inequality for the base case \(n=4\). Calculate both sides of the inequality \(2^{n}\) and \(n !\). For \(n=4\), \(2^{4}=16\) and \(4!=24\). This shows that for \(n=4\), the inequality \(2^{n}<n !\) holds true as \(16<24\).
02

Inductive Step - Assume true for \(k\)

Assume that the inequality holds true for \(n = k\), that is, assume \(2^{k}<k !\).
03

Inductive Step - Prove true for \(k+1\)

To prove the inductive step, show that if \(2^{k} 2\) for \(k \geq 4\), the inequality \(2 * 2^{k} < (k+1) * k!\) is true.
04

Conclusion

Since the inequality is true for the base case and inductive step, the inequality \(2^{n}<n !\) for all \(n \geq 4, n \in \mathbb{N}\) is proved by mathematical induction.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(f(x):=1 / x^{2}, x \neq 0, x \in \mathbb{R}\) (a) Determine the direct image \(f(E)\) where \(E:=\\{x \in \mathbb{R}: 1 \leq x \leq 2\\}\). (b) Determine the inverse image \(f^{-1}(G)\) where \(G:=\\{x \in \mathbb{R}: 1 \leq x \leq 4\\}\).

Prove that \(1 / \sqrt{1}+1 / \sqrt{2}+\cdots+1 / \sqrt{n}>\sqrt{n}\) for all \(n \in \mathbb{N}\).

Exhibit a bijection between \(\mathbb{N}\) and a proper subset of itself.

Find the largest nacural number \(m\) such that \(n^{3}-n\) is divisible by \(m\) for all \(n \in \mathbb{N}\). Prove your assertion.

Show that if \(f: A \rightarrow B\) and \(G, H\) are subsets of \(B\), then \(f^{-1}(G \cup H)=f^{-1}(G) \cup f^{-1}(H)\) and \(f^{-1}(G \cap H)=f^{-1}(G) \cap f^{-1}(H)\).
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.