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Chapter 6: Problem 48

Prove by induction that \(1 \cdot 1 !+2 \cdot 2 !+\cdots+n \cdot n !=(n+1) !-1, n \geq 1\).

Short Answer

Expert verified
We prove the formula \(1 \cdot 1! + 2 \cdot 2! + \cdots + n \cdot n! = (n+1)! - 1\) by induction. 1. Base case (n=1): \(1 \cdot 1! = (1+1)! - 1\), which is true; 2. Inductive hypothesis: Assume the formula is true for n=k; 3. Inductive step: Show the formula for n=k+1 using the inductive hypothesis: \(((k+1)! - 1) + (k+1) \cdot (k+1)!\) \(=(k+1)!(1 + (k+1))\) \(=(k+2)! - 1\). By the principle of mathematical induction, the formula holds for all n \(\geq 1\).

Step by step solution

01

Prove the Base Case (n=1)

First, we need to prove the formula for the base case where n=1. By substituting n=1 in the given formula, we get: \(1 \cdot 1! = (1+1)! - 1\) \(1 \cdot 1 = 2! - 1\) \(1 = 2 - 1\), which is true. So the formula holds for n=1.
02

Inductive Hypothesis

For the inductive step, assume the formula is true for n=k: \(1 \cdot 1! + 2 \cdot 2! + \cdots + k \cdot k! = (k+1)! - 1\)
03

Prove the Formula for n=k+1

Now, we need to show that the formula holds for n=k+1: \(1 \cdot 1! + 2 \cdot 2! + \cdots + k \cdot k! + (k+1) \cdot (k+1)!\) \(=((k+1)! - 1) + (k+1) \cdot (k+1)!\) (by the inductive hypothesis) Now, we need to show that this is equal to \(((k+2)!) - 1\): \(((k+1)! - 1) + (k+1) \cdot (k+1)!\) \(=(k+1)! - 1 + (k+1)! \cdot (k+1)\) \(= (k+1)!(1 + (k+1))\) \(=(k+1)!(k+2)\) Now, we recall that \((k+2)! = (k+2)(k+1)!\), so we have: \(=(k+2)! - 1\) So the formula holds for n=k+1.
04

Conclusion

Since we have shown that the formula is true for the base case (n=1) and for any n=k+1 if it is true for n=k, the formula holds for all n \(\geq 1\) by induction.

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Key Concepts

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

Proof by Induction
Proof by induction is a powerful mathematical technique used to prove that a statement is true for all natural numbers. It consists of two main steps: the base case and the inductive step.

In the base case, you prove that the statement is true for the initial value, usually when n equals 1 or 0. In the example provided, we've confirmed that the given formula holds true when n is 1. Then, the inductive hypothesis assumes the statement holds for some arbitrary positive integer k. This serves as the launching pad for the inductive step, where we demonstrate the statement's validity for k+1. If both the base case and inductive step are verified, we conclude that the statement is true for all natural numbers due to the principle of mathematical induction. This reasoning mirrors a row of dominoes falling: if the first one falls (base case), and each domino makes the next fall over (inductive step), then all the dominoes will fall.
Factorial
The factorial of a non-negative integer n, denoted as n!, is the product of all positive integers less than or equal to n. In a sense, it's like a hyper-accelerated addition because you're multiplying numbers in a sequence together instead of adding them. For instance, 5! = 5 x 4 x 3 x 2 x 1.

In our example, the factorial plays a key role in both the formulation and solution of the problem. Understanding factorials is critical because it defines how mathematical expressions expand or simplify when dealing with sequences, permutations, and combinations in discrete mathematics and beyond.
Inductive Hypothesis
The inductive hypothesis is an assumption made during the second step of proof by induction. It's where we suppose that our statement works perfectly fine when we're at some step k in our sequence of natural numbers. This isn't just wishful thinking; rather, it's a strategic move that sets us up to prove the statement also works for k+1.

For example, in the given exercise, by accepting that our formula holds at n=k, we can confidently step forward to check if this presumption allows the statement to stay true for the next step, which is n=k+1. This logical leap from k to k+1 is akin to a rite of passage that each number must go through for the statement to be universally accepted.
Discrete Mathematics
Discrete mathematics is the study of mathematical structures that are fundamentally discrete rather than continuous. In other words, it deals with countable, separated values rather than smooth and connected values. This branch of mathematics includes topics such as combinatorics, graph theory, and theoretical computer science.

Proof by induction, like in our exercise, is a staple in discrete mathematics because it is perfect for statements about integers, which are discrete by nature. Understanding the principles of discrete mathematics is crucial when diving into algorithms, systematizing data, and much more in the realm of mathematics and computer science.

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Most popular questions from this chapter

A survey of 475 customers at Chestnut Restaurant shows that of the three ice cream flavors - chocolate, strawberry, and vanilla -65 customers like only chocolate, 75 like only strawberry, 85 like only vanilla, 100 like chocolate but not strawberry, 120 like strawberry but not vanilla, 140 like vanilla but not chocolate, and 65 like none of the flavors. Find the probability that a customer selected at random from the survey: Likes chocolate, given that she does not like strawberry or vanilla.

Evaluate each sum. $$ a\left(\begin{array}{l}{n} \\\ {0}\end{array}\right)+(a+d)\left(\begin{array}{l}{n} \\\ {1}\end{array}\right)+(a+2 d)\left(\begin{array}{l}{n} \\\ {2}\end{array}\right)+\cdots+(a+n d)\left(\begin{array}{l}{n} \\\ {n}\end{array}\right) $$ (Hint: Use the same hint as in Exercise \(34 .\) )

The Sealords have three children. Assuming that the outcomes are equally likely and independent, find the probability that they have three boys, knowing that: At least one child is a boy.

The Bell numbers \(B_{n},\) named after the English mathematician Eric T. Bell (1883-1960) and used in combinatorics, are defined recursively as follows: $$B_{0}=1$$ $$B_{n}=\sum_{i=0}^{n-1}\left(\begin{array}{c} n-1 \\ i \end{array}\right) B_{i}, \quad n \geq 1$$ Compute each Bell number. $$B_{3}$$

Using induction, prove each. $$\left(\begin{array}{l}n \\ 0\end{array}\right)^{2}+\left(\begin{array}{l}n \\\ 1\end{array}\right)^{2}+\left(\begin{array}{l}n \\\ 2\end{array}\right)^{2}+\cdots+\left(\begin{array}{l}n \\\ n\end{array}\right)^{2}=\left(\begin{array}{l}2 n \\ n\end{array}\right)$$
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