91Ó°ÊÓ

The base step is the starting point in the process of mathematical induction. It's where we test the given statement to see if it's true for the first positive integer, typically 1. By verifying the statement at this initial step, we establish a foundation that supports the rest of the inductive process.
In the given problem, we start by substituting 1 for \(n\) in the expression \(n^2 + 3n\). Our expression then becomes \(1^2 + 3 \times 1 = 4\). Since 4 is exactly divisible by 2, the statement "2 is a factor of \(n^2 + 3n\)" holds true for \(n = 1\).
This base step acts as our solid foothold, ensuring that we can start the inductive chain with a true premise.

The inductive step is the next phase in mathematical induction. Here we aim to show that if the statement is true for an integer \(n = k\), then it must also be true for \(n = k + 1\).
We begin by assuming the statement holds for \(n = k\), meaning 2 is a factor of \(k^2 + 3k\), or mathematically, \(k^2 + 3k = 2m\) for some integer \(m\). This is our induction hypothesis.

• To complete this step, we substitute \(k + 1\) for \(n\) in the expression \(n^2 + 3n\).
• By doing so, the expression transforms to \((k + 1)^2 + 3(k + 1)\).
• Simplifying it results in \(k^2 + 3k + 2k + 4\), which can be rewritten as \(2(m + 2k + 2)\).

This form again shows 2 as a factor so the assumption holds for \(n = k + 1\). Thus, with both steps validated, the statement stands proven by induction.

A positive integer is simply a whole number greater than zero. Positive integers start from 1 and go upwards without ending, like 1, 2, 3, and so on.
They play a crucial role in mathematical induction since the principle applies to statements about positive integers. In this problem, the solution defends the validity of the statement "2 is a factor of \(n^2 + 3n\)" for every positive integer \(n\).
In other words, regardless of the positive integer you choose for \(n\), the proof aims to show that the expression \(n^2 + 3n\) will always be divisible by 2. This unending applicability to all positive integers is what makes mathematical induction a powerful and flexible proof technique in mathematics.

Factorization involves writing a number or expression as a product of its factors. Factors are numbers you can multiply together to get another number. For example, the factors of 4 are 2 and 2, because \(2 \times 2 = 4\).
In the context of the exercise, factorization is essential in demonstrating the inclusion of the factor 2 in \(n^2 + 3n\). In both the base step and inductive step, we need to show that the entire expression conforms to a form like \(2m\), where 2 explicitly appears as a factor in the mathematical composition.
By rewriting \(k^2 + 3k\) and allowing it to become \(2m\), and similarly refactoring the \(k+1\) substitution to gain \(2(m + 2k + 2)\), we ensure that factor 2 incorporates into the solution. Such manipulations confirm the divisibility or factorization of 2 from these generated expressions, fulfilling the need for proof in this exercise.