Problem 73 - College Algebra

Question

Show that the sum of the first \(n\) positive odd integers, $$i+3+5+\cdots+(2 n-1)$$ is \(n^{2}.\)

Step-by-Step Solution

Answer

The sum of the first \(n\) positive odd integers is indeed \(n^{2}\). This statement has been proven correct using mathematical induction.

Step 1: Setup

The sum \(S\) of the first \(n\) positive odd numbers can be stated as \(S = 1 + 3 + 5 +...+ (2n - 1)\). The task is to show \(S = n^{2}\). This can be proven using mathematical induction.

Step 2: Base Case

For \(n = 1\), the sum \(S\) equals to the first and only term which is 1. Thus, \(S = 1^{2} = 1\), so the statement holds for \(n = 1\).

Step 3: Assumption

Assume the statement is true for \(n = k\), that is \(S_k =1 + 3 + 5 + \cdots + (2k - 1) = k^{2}\).

Step 4: Induction Step

Next, prove the statement holds for \(n = k + 1\), i.e., \(S_{k+1} = 1 + 3 + 5 + \cdots + (2k - 1) + (2(k + 1) - 1) = (k + 1)^{2}\). Substitute the assumption into the equation: \(S_{k+1} = k^{2} + 2(k + 1) - 1 = (k^2 + 2k + 2 -1)\). After simplifying the equation, you will find that \(S_{k+1} = (k^2 + 2k + 1) = (k + 1)^{2}\). This completes the proof.

Key Concepts

Sum of Odd Integers

When you look at the sequence of odd integers like 1, 3, 5, and so forth, an important fact is that their sum up to any number of terms can be represented in an elegant formula. Specifically, the sum of the first \(n\) positive odd integers is \(n^2\). For example, the sum of the first odd number is 1 which equals \(1^2\). The sum of the first two odd numbers is \(1 + 3 = 4\), which equals \(2^2\). You can extend this pattern for any number of terms: \(1 + 3 + 5 + \cdots + (2n - 1) = n^2\).

This concept forms the foundation of many mathematical explorations. It's not just a formula, but a beautiful insight into how numbers work in harmony.

Proof by Induction

Mathematical induction is a powerful technique used to prove statements that are believed to be true for every natural number. The process is akin to proving that a line of dominoes will fall: if we can make sure the first domino falls (base case) and show that if one domino falls the next one will too (inductive step), all the dominoes will fall.

It's particularly useful for propositions that have a clear pattern or progression, such as the sum of odd integers. By establishing both a base case and an inductive step, we can confidently assert that a statement is universally true for all integers in the sequence.

Base Case

In any mathematical induction, the base case is the starting point of the proof. It's the specific instance where we first check that our formula holds true. For our exercise, the base case is when \(n = 1\).

Here, the sum of the first odd integer is 1, and the expected result given by \(n^2\) is also 1. As these two match, our base case confirms that the formula is correct for \(n = 1\). This crucial first step acts like the first domino which ensures the entire sequence is valid.

Inductive Step

The inductive step is the heart of the mathematical induction proof. After establishing the base case, the inductive step involves assuming that the statement is true for some arbitrary number \(n = k\). You then demonstrate that if it holds for \(n = k\), it must also hold for \(n = k + 1\).

In this exercise, we assume \(S_k = k^2\), which is our assumption that the sum of the first \(k\) odd numbers is \(k^2\). We then add the next odd integer, \(2(k + 1) - 1\), to ensure the formula remains true for \(k + 1\). After simplifying, you will find \(S_{k+1} = (k+1)^2\). This part of the proof confirms the domino effect: one true case leads to the next, covering all natural numbers.