Problem 8 - Algebra 2

Question

Prove that each statement is true for all positive integers. $$ 1+5+9+\cdots+(4 n-3)=n(2 n-1) $$

Step-by-Step Solution

Answer

Proved by induction, the statement is true for all positive integers.

Step 1: Understand the Problem

We need to show that the sum of the sequence \(1 + 5 + 9 + \cdots + (4n - 3)\) is equal to \(n(2n - 1)\) for all positive integers \(n\).

Step 2: Base Case

First, let's verify the statement for the base case with \(n = 1\). The sequence has only one term, \((4 \times 1 - 3) = 1\), and the right side is \(1(2 \times 1 - 1) = 1\). Thus, the base case holds as both sides are equal to 1.

Step 3: Inductive Hypothesis

Assume that for a positive integer \(k\), the statement holds true: \(1 + 5 + 9 + \cdots + (4k - 3) = k(2k - 1)\). This is our inductive hypothesis.

Step 4: Inductive Step

We need to show the statement is true for \(n = k + 1\). The left-hand side becomes \(1 + 5 + 9 + \cdots + (4k + 1)\). We can split this as \([1 + 5 + 9 + \cdots + (4k - 3)] + (4(k + 1) - 3)\).

Step 5: Simplify the Sum

Using the inductive hypothesis, the sum \(1 + 5 + 9 + \cdots + (4k - 3) = k(2k - 1)\). Adding the next term gives: \(k(2k - 1) + (4k + 1)\).

Step 6: Simplify the Right Side

Calculate \((k + 1)(2(k + 1) - 1)\). This simplifies as \((k + 1)(2k + 2 - 1) = (k + 1)(2k + 1)\).

Step 7: Verification

Verify that \(k(2k - 1) + (4k + 1) = (k + 1)(2k + 1)\). Expanding the right side gives \(2k^2 + k\). Expanding the left side \(k(2k - 1) + 4k + 1 = 2k^2 - k + 4k + 1\) simplifies to \(2k^2 + 3k + 1\), which equals \((k + 1)(2k + 1)\). Thus, both sides match.

Step 8: Conclusion

The base case holds, and the inductive step verifies the formula for \(n = k+1\) assuming it holds for \(n = k\). By mathematical induction, the statement is true for all positive integers.

Key Concepts

Algebra

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating those symbols. In our current context, algebra plays a crucial role in simplifying and proving equations. When dealing with algebraic expressions, we often substitute numerical values to verify equations or transform expressions to demonstrate their equality.

• The expression \(1 + 5 + 9 + \cdots + (4n-3)\) is known as an arithmetic expression where each term increases by a constant difference.
• The algebraic equation \(n(2n-1)\) helps summarize and generalize the sequence's pattern.
• A common task in algebra is to equate expressions on both sides of an equation, leading to simplified forms or proofs.

Algebra simplifies real-world problems by transforming them into solvable equations.

Sequence

A sequence, in mathematics, is an ordered list of numbers. In our problem, the sequence is \(1, 5, 9, \ldots, (4n-3)\). Each number in a sequence is called a term.
The sequence in this problem has some distinct characteristics:

• It is an arithmetic sequence, meaning each term increases by the same value, which we call the common difference.
• For this specific sequence, the common difference between terms is 4, as seen from \(5 - 1 = 4\) and \(9 - 5 = 4\).
• The formula for the general term of this sequence is \(4n - 3\), defining each term in relation to its position \(n\) in the sequence.

Understanding and using sequences helps in modeling patterns and predicting future terms.

Series

A series is the sum of the terms of a sequence. In the exercise, the series is expressed as \(1 + 5 + 9 + \cdots + (4n-3)\). The terms are added to produce a single sum.
When dealing with series:

• Each term’s contribution to the total is considered, leading to the generalized result \(n(2n-1)\).
• The series we examine is finite, meaning it ends after a certain number of terms.
• Finite series often have a formula, as we have here, that allows us to calculate the sum without adding each term individually.

Recognizing and working with series is critical in various fields, from calculating financial investments to approximating complex scientific models.

Proof by Induction

Proof by induction is a powerful mathematical technique often used to prove statements that are asserted to be true for all positive integers. The method involves several steps:

• Base Case: Start by proving that the statement holds for the first integer (often \(n = 1\)). In this problem, we verified the base case where the single term of the series equals the right-hand side of the equation.
• Inductive Hypothesis: Assume the statement is true for an arbitrary positive integer \(k\). This is the step of assuming a general case is true to build upon for the next step.
• Inductive Step: Demonstrate that if the statement holds for \(k\), it must also hold for \(k+1\). Simplifying both sides and showing their equality is crucial here.

Effectively using proof by induction helps strengthen logical reasoning and is a foundational technique in algebra and other mathematical areas.