Problem 17 - Algebra and Trigonometry

Question

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+n(n+1)=\frac{1}{3} n(n+1)(n+2) $$

Step-by-Step Solution

Answer

True for all natural numbers \( n \).

Step 1 - Base Case

Verify the statement for the smallest natural number, which is typically 1. For\( n = 1 \): 1) Left-hand side (LHS): \( 1 \times 2 = 2 \), 2) Right-hand side (RHS): \( \frac{1}{3} \times 1 \times (1+1) \times (1+2) = \frac{1}{3} \times 1 \times 2 \times 3 = 2 \) Since the LHS equals the RHS, the base case holds true.

Step 2 - Inductive Hypothesis

Assume the statement is true for some natural number \( k \). That is, assume \[ 1 \times 2 + 2 \times 3 + 3 \times 4 + \cdots + k(k+1) = \frac{1}{3} k(k+1)(k+2). \]

Step 3 - Inductive Step

Show the statement holds for \( k+1 \). That is, prove \[ 1 \times 2 + 2 \times 3 + 3 \times 4 + \cdots + k(k+1) + (k+1)(k+2) = \frac{1}{3} (k+1)(k+2)(k+3). \] Using the inductive hypothesis, the left-hand side becomes \[ \frac{1}{3} k(k+1)(k+2) + (k+1)(k+2). \] Factor out \( (k+1)(k+2) \) from the sum: \[ (k+1)(k+2) \left( \frac{k}{3} + 1 \right) = (k+1)(k+2) \left( \frac{k+3}{3} \right) = \frac{1}{3} (k+1)(k+2)(k+3). \] This matches the desired right-hand side expression, so the statement holds for \( k+1 \).

Step 4 - Conclusion

By the Principle of Mathematical Induction, the statement is true for all natural numbers \( n \).

Key Concepts

Base Case

In mathematical induction, the **Base Case** is the starting point. We begin by verifying the given statement for the smallest natural number, typically 1. This helps us understand if the formula or statement we are trying to prove is initially correct.

In our problem, we check the base case for **n = 1**.

• Left-hand side (LHS): The formula simplifies to \(1 \times 2 = 2\).
• Right-hand side (RHS): The formula simplifies to \[ \frac{1}{3} \times 1 \times (1+1) \times (1+2) = \frac{1}{3} \times 1 \times 2 \times 3 = 2 \].

Since the LHS equals the RHS, this confirms that our base case is true and the statement works for **n = 1**.

Inductive Hypothesis

The next phase of mathematical induction involves the **Inductive Hypothesis**. Here we assume that the statement holds true for some arbitrary natural number **k**. This assumption allows us to use this hypothetical validity to show that the statement also holds for **k+1**.

In our problem, we assume that the formula is true for **k**. Thus, we assume that \[1 \times 2 + 2 \times 3 + 3 \times 4 + \cdots + k(k+1) = \frac{1}{3} k(k+1)(k+2) \] is a true statement. This assumption doesn't prove that it's true for all natural numbers yet; it merely sets the stage for the next step.

Inductive Step

The **Inductive Step** is the crux of mathematical induction. Here, we show that if the statement holds for some arbitrary natural number **k**, it also holds for **k+1**. This step effectively links our base case to all subsequent numbers.

Using our assumption from the inductive hypothesis, we need to show that the statement holds for **k+1**. We start from our assumed statement and add the next term:

• Left-hand side (LHS): \[ 1 \times 2 + 2 \times 3 + 3 \times 4 + \cdots + k(k+1) + (k+1)(k+2) = \frac{1}{3} k(k+1)(k+2) + (k+1)(k+2) \]
• Right-hand side (RHS): We want to show it equals \[ \frac{1}{3} (k+1)(k+2)(k+3) \].

We factor out \( (k+1)(k+2) \) from the lhs sum and simplify:

• Factor: \[ (k+1)(k+2) \left( \frac{k}{3} + 1 \right) \]
• Simplify: \[ (k+1)(k+2) \left( \frac{k+3}{3} \right) = \frac{1}{3} (k+1)(k+2)(k+3) \]

Our simplified expression matches the rhs, hence proving the step and holding the statement true for **k+1**.

Natural Numbers

Mathematical induction often deals specifically with the set of **Natural Numbers**. These numbers are the positive integers starting from 1, 2, 3, and so forth. They are crucial because induction relies on their well-ordered nature and finite increments.

In this problem, we aim to prove the given formula for all **natural numbers n**. Starting with **n = 1** in the **base case**, and assuming it is true for a general number **k** in the **inductive hypothesis**, we then demonstrate it is also true for **k+1** in the **inductive step**.

Since we have shown our base case, hypothesis, and step correctly, by the principle of mathematical induction, the initial statement \[1 \times 2 + 2 \times 3 + 3 \times 4 + \cdots + n(n+1) = \frac{1}{3} n(n+1)(n+2)\] holds true for all natural numbers **n**.

This encapsulates the beauty and utility of mathematical induction, providing a systematic method to prove properties across all natural numbers.