/*! 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 6 Use mathematical induction in Ex... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 6

Use mathematical induction in Exercises \(3-17\) to prove summation formulae. Be sure to identify where you use the inductive hypothesis. Prove that \(1 \cdot 1 !+2 \cdot 2 !+\cdots+n \cdot n !=(n+1) !-1\) whenever \(n\) is a positive integer.

Short Answer

Expert verified
Proved by induction that \(1 \times 1! + 2 \times 2! + \cdots + n \times n! = (n + 1)! - 1 \).

Step by step solution

01

Base Case

To prove by induction, start with the base case where \( n = 1 \). Substitute \( n = 1 \) into the formula: \( 1 \times 1! = (1+1)! - 1 \). This simplifies to \( 1 \times 1 = 2! - 1 \). Since \( 2! = 2 \), we get \( 1 = 1 \), which is true. Thus, the base case holds.
02

Inductive Hypothesis

Assume the statement is true for some positive integer \( k \). That is, \( 1 \times 1! + 2 \times 2! + \cdots + k \times k! = (k + 1)! - 1 \). This is the inductive hypothesis.
03

Inductive Step

We need to prove the statement for \( k + 1 \). Consider the left-hand side for \( n = k + 1 \): \( 1 \times 1! + 2 \times 2! + \cdots + k \times k! + (k + 1) \times (k + 1)! \). By the inductive hypothesis, replace the first part: \( (k + 1)! - 1 + (k + 1) \times (k + 1)! \). Factor \( (k + 1)! \) out: \( (k + 1)! [1 + (k + 1)] - 1 = (k + 2)! - 1 \).
04

Conclusion

Since the statement holds for \( k \) and we have shown it holds for \( k + 1 \), it follows by mathematical induction that \( 1 \times 1! + 2 \times 2! + \cdots + n \times n! = (n + 1)! - 1 \) for all positive integers \( n \).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Summation Formula
In mathematics, a summation formula lets us sum up a series of terms in a compact way. For example, the given problem asks us to sum up the series: \(1 \cdot 1! + 2 \cdot 2! + \cdots + n \cdot n!\).
Let鈥檚 understand **summation** by breaking it down:
  • **Summation** is the process of adding a series of numbers, often represented by the Greek letter sigma (\(\Sigma\)).
  • We'll use this notation: \( \sum_{i=1}^{n} i \cdot i! \), where each term \(i \cdot i!\) is added from \(i=1\) to \(i=n\).
In our problem, it is given that the sum equals \((n+1)! - 1\). This means when we add the products of numbers and their factorials from 1 to \(n\), we get the factorial of \((n+1)\) minus 1.
Factorial
**Factorials** are a foundational concept in mathematics. A factorial of a non-negative integer \(n\) is the product of all positive integers less than or equal to \(n\).
  • It鈥檚 denoted by \(n!\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
  • By convention, \(0! = 1\).
Factorials grow very quickly as \(n\) increases. In our exercise, the formula involves terms like \(i \cdot i!\).
As an example:
  • For \(n = 3\), the sum is \(1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! = 1 + 4 + 18 = 23\), which equals \(4! - 1\).
This rapid growth in factorial values is key to understanding the formula \((n+1)! - 1\).
Inductive Hypothesis
The **inductive hypothesis** is the assumption step in mathematical induction. It helps us extend the truth from one case to the next.
  • We assume our statement is true for a particular positive integer \(k\), like saying, *If this works for some number \(k\), it should also work for \(k+1\).*
  • In symbols, the inductive hypothesis for our formula is: \(1 \cdot 1! + 2 \cdot 2! + \cdots + k \cdot k! = (k + 1)! - 1\).
Next, we'll use this assumption to prove the statement for the next integer, \(k+1\). This step ensures the continuity of the formula for all integers.
Base Case
Every induction problem starts with a **base case**. The base case verifies that the statement is true for the initial value of \(n\).
  • For our exercise, we start with \(n = 1\).
  • Substituting \(n = 1\) into the formula gives: \(1 \cdot 1! = (1+1)! - 1 \)
  • This simplifies to: \(1 = 2! - 1\) or \(1 = 1\), a true statement.
Once the base case is verified, we can confidently use induction to prove the statement for all positive integers, building a strong foundation for the argument.

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

Devise a recursive algorithm to find the \(n\) th term of the sequence defined by \(a_{0}=1, a_{1}=2,\) and \(a_{n}=a_{n-1} \cdot a_{n-2}\) for \(n=2,3,4, \ldots\)

Deal with values of iterated functions. Suppose that \(f(n)\) is a function from the set of real numbers, or positive real numbers, or some other set of real numbers, to the set of real numbers such that \(f(n)\) is monotonically increasing [that is, \(f(n)< f(m)\) when \(n< m )\) and \(f(n)< n\) for all \(n\) in the domain of \(f . ]\) The function \(f^{(k)}(n)\) is defined recursively by $$f^{(k)}(n)=\left\\{\begin{array}{ll}{n} & {\text { if } k=0} \\\ {f\left(f^{(k-1)}(n)\right)} & {\text { if } k>0}\end{array}\right.$$ Furthermore, let \(c\) be a positive real number. The iterated function \(f_{c}^{*}\) is the number of iterations of \(f\) required to reduce its argument to \(c\) or less, so \(f_{c}^{*}(n)\) is the smallest nonnegative integer \(k\) such that \(f^{k}(n) \leq c\). Let \(f(n)=\sqrt{n} .\) Find a formula for \(f^{(k)}(n) .\) What is the value of \(f_{2}^{*}(n)\) when \(n\) is a positive integer?

Find the flaw with the following "proof" that \(a^{n}=1\) for all nonnegative integers \(n,\) whenever \(a\) is a nonzero real number. Basis Step: \(a^{0}=1\) is true by the definition of \(a^{0}\) . Inductive Step: Assume that \(a^{j}=1\) for all nonnegative integers \(j\) with \(j \leq k\) . Then note that $$ a^{k+1}=\frac{a^{k} \cdot a^{k}}{a^{k-1}}=\frac{1 \cdot 1}{1}=1 $$

The well-ordering property can be used to show that there is a unique greatest common divisor of two positive integers. Let \(a\) and \(b\) be positive integers, and let \(S\) be the set of positive integers of the form \(a s+b t,\) where \(s\) and \(t\) are integers. a) Show that \(S\) is nonempty. b) Use the well-ordering property to show that \(S\) has a smallest element \(c .\) c) Show that if \(d\) is a common divisor of \(a\) and \(b\) , then \(d\) is a divisor of \(c .\) d) Show that \(c | a\) and \(c | b .\) [Hint: First, assume that \(c / a .\) Then \(a=q c+r,\) where \(0

Show that the principle of mathematical induction and strong induction are equivalent; that is, each can be shown to be valid from the other.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Logic and Functions

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.