Problem 21 - College Algebra

Question

In Exercises 11-24, use mathematical induction to prove that each statement is true for every positive integer \(n.\) $$1 \cdot 2+2 \cdot 3+3 \cdot 4+\dots+n(n+1)=\frac{n(n+1)(n+2)}{3}$$

Step-by-Step Solution

Answer

Through Mathematical Induction, the given formula has been proven to be true for all positive integers. The Base Case was proven to be true for \(n=1\). Assuming the validity of the expression for \(n=k\) (Inductive Hypothesis), it was demonstrated that it holds for \(n=k+1\) (Inductive Step). Thus, the formula \(1 \cdot 2+2 \cdot 3+3 \cdot 4+\dots+n(n+1)=\frac{n(n+1)(n+2)}{3}\) is true for all positive integer values of \(n\).

Step 1: Base Case

Check if the statement is indeed true when \(n=1\). Plug into the equation \(1 \cdot 2\), which equals 2. On the other hand, calculate the right side of the equation \[\frac{1(1+1)(1+2)}{3}=2\]. Both sides equal, which means that our statement holds for \(n = 1\).

Step 2: Inductive Hypothesis

Let's now make the inductive hypothesis. This step involves assuming that the statement is true for \(n=k\), where \(k\) is some positive integer. \[1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + k(k+1) = \frac{k(k+1)(k+2)}{3}\]

Step 3: Inductive Step

Use the induction hypothesis to show the statement is true for \(n=k+1\). For \(n=k+1\), the left side of the equation becomes: \[1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + k(k+1) + (k+1)(k+2).\] Assume that from the inductive hypothesis that the sum of the first terms is \[\frac{k(k+1)(k+2)}{3}\], the left side of the equation becomes \[\frac{k(k+1)(k+2)}{3} + (k+1)(k+2).\] Now try to prove that this is equal to the other side for \(n=k+1\), a.k.a. \[\frac{(k+1)((k+1)+1)((k+1)+2)}{3} = \frac{(k+1)(k+2)(k+3)}{3}.\] If these two forms are equal, then we confirm that our statement is indeed true for every positive integer number \(n\).

Step 4: Proving the Inductive Step

Firstly factor out \((k+1)(k+2)/3\) on the left hand side, which results in \[\frac{(k+1)(k+2)}{3}(k+3).\] Then compare to the right hand side of the equation, which is also \[\frac{(k+1)(k+2)(k+3)}{3}.\] As the left side is equal to the right side, we have completed the proof via mathematical induction. Thus, the statement is true for all positive integers.

Key Concepts

Proof by Induction

Proof by induction is a powerful mathematical technique used to prove that a given statement holds true for all positive integers. The process involves two steps: establishing a base case and performing an inductive step.

The base case typically examines the simplest scenario, often when the integer value is 1, and confirms that the statement is true at this starting point. As demonstrated in the original exercise, when setting n to 1, both sides of the given equation yield the same result, thus confirming the base case.

Following a successful base case, the inductive step requires the assumption that the statement is true for some integer k (inductive hypothesis). Then, the proof shows that if the statement holds for k, it must also be true for k+1. This logical progression from k to k+1 is the essence of the inductive step, effectively 'stepping forward' and bridging each integer with its successor to cover all positive integers.

Inductive Hypothesis

The inductive hypothesis is the assumption within proof by induction that the statement we're trying to prove is true for some arbitrary positive integer k. This assumed truth acts as a pivot point for proving the statement's truth for the next integer in line, k+1.

In our given exercise, the inductive hypothesis is that the sum of the series 1⋅2 + 2⋅3 + 3⋅4 + ... + k(k+1) equals \(\frac{k(k+1)(k+2)}{3}\). This hypothesis is not proven yet; instead, it is assumed to be true for the purposes of conducting the inductive step. It allows for extrapolation under the premise that if it works for k, under the logical structure of the problem, it must also work for k+1, thereby chaining the truth from one integer to the next.

Sequences and Series

Sequences and series are foundational concepts in mathematics, particularly in algebra and calculus.

A sequence is an ordered list of numbers, where each number is called a term. For example, the sequence 1, 2, 3, 4, ... is a simple arithmetic sequence where each term is one more than the previous term.

A series is the sum of the terms of a sequence. If the sequence is finite, the sum is called a finite series, and if the sequence is infinite, the sum is called an infinite series, provided it converges to a specific value.

In the exercise provided, we are dealing with a finite series of the form 1⋅2 + 2⋅3 + 3⋅4 + ... + n(n+1). Understanding the nature of this series is key to creating and verifying a formula that could help to sum the series efficiently, which is precisely the goal of employing mathematical induction to find a generalized formula.

Algebraic Proof

An algebraic proof is a type of proof in mathematics that relies on algebraic techniques to demonstrate the validity of a statement. In our example, the induction process involves manipulating algebraic expressions to show that both sides of an equation are equivalent.

The algebraic manipulation in the inductive step typically rearranges terms, factors expressions, and simplifies the algebra to reach a form that matches the assumed truth for k+1. This process may include common steps like factoring out common terms, as shown in the final proving step of our problem.

In our exercise, the left side of the equation is modified by factoring, which makes it visibly equal to the right side when we assume the induction hypothesis is true for k+1. The simplification that leads to the conclusion validates the inductive step of the proof and hence, confirms the entire algebraic proof by induction of the original statement for all positive integers.