Problem 22

Question

Use mathematical induction to prove that each statement is true for every positive integer n. $1 \cdot 3+2 \cdot 4+3 \cdot 5+\cdots+n(n+2)=\frac{n(n+1)(2 n+7)}{6}$

Step-by-Step Solution

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Answer

The given statement is not true for all positive integers as it fails for $n = 1$, which negates the base case for induction. Therefore, we cannot apply mathematical induction to prove the given statement.

1Step 1: Basis of Induction

The base case for mathematical induction is to verify the statement for $n = 1$. So let's plug in $n = 1$ into both sides of the equation. The left hand side becomes $1 \cdot 3 = 3$, and the right hand side is $\frac{1(1+1)(2 \cdot 1+7)}{6} = \frac{10}{6} = \frac{5}{3}$. The statement does not hold for $n = 1$. Therefore, this problem is impossible to solve, as the base case of induction is false.

2Step 2: Inductive Step

Given that the base case for induction is false, the inductive step is irrelevant and does not need to be checked. In general, the principle of mathematical induction states that if the statement holds for all $ n = k $, then it should also hold for $ n = k + 1 $. However, since the base case of our induction (n=1) doesn't hold true in this case, hence we cannot proceed with the induction step.

Key Concepts

Base CaseInductive StepPositive IntegerProof by Induction

Base Case

The base case is the starting point in the method of mathematical induction. When using induction, the first step is to verify that the given statement is true for the initial value, often noted as $ n = 1 $. This verification forms the foundation for the entire proof. The base case is crucial because:

It establishes that the statement holds true for at least one instance.
It provides the basis from which all further steps are built.

In the provided exercise, when we attempted to verify the base case for $ n = 1 $, the statement did not hold true. Specifically, the left side of the equation did not equal the right side. If the base case fails, as in this situation, we cannot move forward with the proof, and the statement is not valid for all positive integers.

Inductive Step

In a mathematical induction proof, the inductive step is the process used to show that if the statement holds for a particular positive integer $ n = k $, then it must also be true for the next integer $ n = k + 1 $. This step is vital because:

It helps to establish a chain of truth, connecting each integer to its successor.
If successful, it illustrates that the formula is valid not just for one instance but across all subsequent positive integers starting after the base case.

During the inductive step, you assume that the formula is true for $ n = k $, known as the induction hypothesis. Then, demonstrate it for $ n = k + 1 $. In this exercise, however, we skipped the inductive step because the base case failed to hold true.

Positive Integer

A positive integer is any whole number greater than zero. These numbers are crucial in mathematical proofs such as induction because they serve as the domain over which statements are tested. For example, positive integers include:

1
2
3
4
And so on...

When using mathematical induction, we start with the smallest positive integer (usually 1) as our base case. From there, assuming the base case holds, we prove the statement step by step for each successive integer. Ineffectively addressing the base case with these integers can derail an induction proof, as seen in the original exercise.

Proof by Induction

Proof by induction is a powerful technique used to prove that a statement is true for all positive integers. It involves two key steps: the base case and the inductive step. When successfully applied, this method can succinctly demonstrate the validity of propositions across the infinite set of positive integers.Here's how proof by induction works:

Begin with the base case, which involves proving that the statement is true for the smallest positive integer, typically $ n=1 $.
Proceed with the inductive step, where you assume the statement is true for $ n=k $ and then prove it for $ n=k+1 $.

If both steps are proven, the statement is true for all positive integers. In the given exercise, the base case did not hold, which invalidated the proof. This highlights the necessity of a true base case to ensure the integrity of proof by induction.

Problem 22

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