Chapter 2: Problem 9
Use induction to prove that \(1+2+2^{2}+\cdots+2^{n}=2^{n+1}-1\) for \(n \in \mathbb{N}\).
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! 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}
Learning Materials
Features
Discover
Chapter 2: Problem 9
Use induction to prove that \(1+2+2^{2}+\cdots+2^{n}=2^{n+1}-1\) for \(n \in \mathbb{N}\).
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
Let \(a\) and \(b\) be integers such that \(\operatorname{gcd}(a, b)=1\). Let \(r\) and \(s\) be integers such that \(a r+b s=1 .\) Prove that $$ \operatorname{gcd}(a, s)=\operatorname{gcd}(r, b)=\operatorname{gcd}(r, s)=1 $$
Show that the Principle of Well-Ordering for the natural numbers implies that 1 is the smallest natural number. Use this result to show that the Principle of Well-Ordering implies the Principle of Mathematical Induction; that is, show that if \(S \subset \mathbb{N}\) such that \(1 \in S\) and \(n+1 \in S\) whenever \(n \in S\), then \(S=\mathbb{N}\).
. Let \(n \in \mathbb{N}\). Use the division algorithm to prove that every
integer is congruent mod \(n\) to precisely one of the integers \(0,1, \ldots,
n-1 .\) Conclude that if \(r\) is an integer, then there is exactly one \(s\) in
\(\mathbb{Z}\) such that \(0 \leq s
Using the division algorithm, show that every perfect square is of the form \(4 k\) or \(4 k+1\) for some nonnegative integer \(k\).
Prove that \(10^{n+1}+10^{n}+1\) is divisible by 3 for \(n \in \mathbb{N}\).
What do you think about this solution?
We value your feedback to improve our textbook solutions.