Problem 1 - Algebra and Trigonometry

Question

Use mathematical induction to prove that the formula is true for all natural numbers n. $$2+4+6+\cdots+2 n=n(n+1)$$

Step-by-Step Solution

Answer

The formula is true for all natural numbers \(n\) by mathematical induction.

Step 1: Understand the Problem

We need to prove that the sum of the first \(n\) even numbers can be expressed using the formula \(2+4+6+\cdots+2n = n(n+1)\) for all natural numbers \(n\). We'll use mathematical induction to do this.

Step 2: Base Case

Check if the formula works for \(n = 1\). Substitute \(n = 1\) into the given formula: \(2 = 1(1+1)\). This simplifies to \(2 = 2\), so the base case is true.

Step 3: Inductive Hypothesis

Assume that the formula is true for some arbitrary natural number \(k\). This means \(2 + 4 + 6 + \cdots + 2k = k(k+1)\) holds true.

Step 4: Inductive Step

We need to prove that if the formula is true for \(k\), then it is also true for \(k+1\). Start with the sum for \(k+1\) terms: \(2 + 4 + 6 + \cdots + 2k + 2(k+1)\).

Step 5: Simplify the Inductive Step Expression

Using the inductive hypothesis, replace the sum \(2 + 4 + 6 + \cdots + 2k\) with \(k(k+1)\). Then, the expression becomes \(k(k+1) + 2(k+1)\).

Step 6: Factor and Simplify Further

Factor \((k+1)\) out of the expression: \(k(k+1) + 2(k+1) = (k+1)(k + 2)\), which simplifies to \((k+1)((k+1) + 1)\). This matches the formula \((k+1)((k+1)+1)\), completing the inductive step.

Step 7: Conclude the Proof

Since the base case is true and the inductive step has been verified, by mathematical induction, the formula \(2 + 4 + 6 + \cdots + 2n = n(n+1)\) is true for all natural numbers \(n\).

Key Concepts

Sum of Even Numbers

Even numbers are numbers that can be exactly divided by 2, leaving no remainder. An interesting property of even numbers arises when we add them in sequence. Consider the simple sequence: 2, 4, 6, and so on. These numbers are consecutive even numbers starting from 2.

• The sum of these first few even numbers takes the form \(2 + 4 + 6 + \cdots + 2n\).
• This sequence can be simplified and expressed with an elegant formula: \(n(n+1)\), where \(n\) represents how many even numbers are included.
• For instance, if \(n = 3\), the sum is \(2 + 4 + 6 = 12\) and \(3 \times (3+1) = 12\), confirming our formula.

Recognizing this formula allows us to quickly find the sum of any collection of consecutive even numbers without manually adding each term.

Algebraic Proof

Algebraic proofs provide a systematic way to verify the truth of a given mathematical formula or statement. In the exercise of proving the sum of even numbers, we are tasked with demonstrating an algebraic identity using the formula \(n(n+1)\).

• An algebraic proof involves starting with an assumption or equation and applying valid algebraic operations to reach a conclusion.
• The key is to manipulate expressions and equations logically while maintaining equality throughout the process.
• This often involves simplification, factoring, and substitution techniques.

By using algebraic properties, we explicitly show that our formula holds true for all integers we are interested in. This verification strengthens our understanding of the relationship between sequences and their algebraic representations.

Base Case

The base case in mathematical induction serves as the foundational step to prove a statement. It is analogous to setting the first domino in a row that leads to a chain reaction. Without a solid first step, the rest of the proof may not stand.

• In this exercise, we begin by establishing the truth of the formula for the smallest natural number, \(n = 1\).
• By direct substitution, we find that the sum for one even number is \(2\), matching the formula \(1(1+1) = 2\).
• This verification affirms that the formula correctly calculates the sum when \(n = 1\), supporting our progression to subsequent steps.

Verifying the base case sets up our entire proof, ensuring the formula's correctness from the very beginning.

Inductive Step

The inductive step is the process where we demonstrate that if the given statement is true for an arbitrary case \(k\), it must also be true for the next case, \(k+1\).

• Our task is to show that if \(2 + 4 + 6 + \cdots + 2k = k(k+1)\), then \(2 + 4 + 6 + \cdots + 2(k+1) = (k+1)((k+1) + 1)\).
• We start by rewriting the sum for \(k+1\) as \(2 + 4 + 6 + \cdots + 2k + 2(k+1)\), then use our inductive hypothesis: \(2 + 4 + 6 + \cdots + 2k = k(k+1)\).
• This leads to \(k(k+1) + 2(k+1) = (k+1)(k+2)\).

By simplifying, we arrive at \((k+1)((k+1)+1)\), which mirrors our original formula for \(n = k+1\). Completing the inductive step confirms the formula holds for every natural number after 1.