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Chapter 10: Problem 17

In Exercises \(5-25,\) prove the statement by induction. $$\begin{aligned} &1 \cdot 2+2 \cdot 3+3 \cdot 4+\dots+n(n+1)\\\ &=\frac{n(n+1)(n+2)}{3} \end{aligned}$$

Short Answer

Expert verified
This completes the process of mathematical induction, and thus it is proved that \(1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ... + n(n+1) = \frac{n(n+1)(n+2)}{3}\) for all positive integers \(n\).

Step by step solution

01

Base Case

Firstly, check if the formula holds for the first case, \(n = 1\). On the left side of the equation, if we put \(n = 1\), we have \(1 \cdot 2 = 2\). On the right side, if we put \(n = 1\) in our formula we get \(\frac{1(1+1)(1+2)}{3} = 2\). So for \(n = 1\), both sides equal, satisfying the base case.
02

Induction Hypothesis

Assume the formula holds true for \(n = k\), where \(k\) is any positive integer. That is, it should be true that \(1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ... + k(k+1) = \frac{k(k+1)(k+2)}{3}\). This is the induction hypothesis.
03

Inductive Step

The next step is to show that the formula holds for \(n = k+1\), if it holds for \(n = k\). Consider the left side of the formula after \(n = k+1\): \(1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ... + k(k+1) + (k+1)((k+1)+1)\). By the inductive hypothesis, \(1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ... + k(k+1)\) can be replaced by \(\frac{k(k+1)(k+2)}{3}\), and thus this becomes \(\frac{k(k+1)(k+2)}{3} + (k+1)(k+2)\). This can be simplified to \(\frac{k(k+1)(k+2)}{3} + \frac{3(k+1)(k+2)}{3} = \frac{(k+1)(k+2)(k+3)}{3}\), which is equal to the right side of the equation when \(n = k+1\). Therefore, the inductive step is also met.

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Key Concepts

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

Inductive Proof
An inductive proof is a powerful mathematical technique often used to show that a given property holds for all natural numbers. The process comprises two main components, the base case and the inductive step, each playing a critical role in establishing the proof's validity.

To understand the value of an inductive proof, picture it as a domino effect. Once the first domino (the base case) falls, it sets off a chain reaction that knocks down every subsequent domino (subsequent values of n). This vivid analogy helps clarify why an inductive proof works: if we can initiate the chain reaction and demonstrate that the effect continues uninterrupted (the inductive step), we can be confident that all dominoes, akin to all natural numbers, will follow suit.
Inductive Hypothesis
The inductive hypothesis forms the pivotal plank in the bridge of mathematical induction. It can be likened to the hypothesis of an experiment; an assumption we make under controlled conditions before verifying it. In the context of our given exercise, the induction hypothesis is not a leap of faith, but rather a logical placeholder. We assume that the formula holds for a particular case n=k, and our task is to prove that because it holds at this step, it will hold for the next one, n=k+1.

Embracing the inductive hypothesis is accepting that we have a working theory, without yet having tested it beyond our base case. It is the powerful 'if' statement in our logical argument--if this, then that--and conquering this step is essential for the proof to continue.
Inductive Step
The inductive step is the critical moment where we leap from the known to the unknown, extending our induction hypothesis to the next natural number. In simpler terms, it is where we show that if the statement is true for n=k, it is also true for n=k+1. This part of the proof is like adding another span to a bridge, ensuring a smooth path over the water; one misstep and the entire argument could collapse.

For our problem, the inductive step involves taking our assumption that the formula works for an arbitrary but specific case of n=k and stretching it to fit n=k+1. By algebraically manipulating the expression, we aim to demonstrate that the pattern not only holds but continues organically, cementing the fact that the initial property we're proving applies universally to all natural numbers following our base case.

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Most popular questions from this chapter

A password for a computer system consists of six characters. Each character must be a digit or a letter of the alphabet. Assume that passwords are not case-sensitive. How many passwords are possible? How many passwords are possible if a password must contain at least one digit? (Hint for second part: How many passwords are there containing just letters?)

State whether the sequence is arithmetic or geometric. $$4,10,16,22, \dots$$

Recreation The following table gives the amount of money, in billions of dollars, spent on recreation in the United States from 1999 to \(2002 .\) (Source: Bureau of Economic Analysis) $$\begin{aligned} &\text { Year } \quad 1999 \quad 2000 \quad 2001 \quad 2002\\\ &\begin{array}{l} \text { Amount } \\ \text { (S billions) } 546.1 \quad 585.7 \quad 603.4 \quad 633.9 \end{array} \end{aligned}$$ Assume that this sequence of expenditures approximates an arithmetic sequence. (a) If \(n\) represents the number of years since 1999 , use the linear regression capabilities of your graphing calculator to find a function of the form \(f(n)=a_{0}+n d, n=0,1,2,3, \ldots,\) that models these expenditures. (b) Use your model to project the amount spent on recreation in 2007

Use counting principles from Section 10.4 to calculate the number of outcomes. A group of friends, five girls and five boys, wants to go to the movies on Friday night. The friends select, at random, two of their group to go to the ticket office to purchase the tickets. What is the probability that the two selected are both boys?

Suppose five cards are drawn from a standard deck of 52 cards. Find the probability that all five cards are black. (Hint: Use the counting principles from Section \(10.4 .)\)
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