Problem 21 - Algebra and Trigonometry

Question

Use mathematical induction to prove that each statement is true for every positive integer n. \(1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+n(n+1)=\frac{n(n+1)(n+2)}{3}\)

Step-by-Step Solution

Answer

The given mathematical expression is true for all positive integers using mathematical induction.

Step 1: Base Case

First, verify the base case. Here the base case is when n=1. Substitute n=1 into the left and right side of given equation. If both sides of equation are equal, then the base case holds true.

Step 2: Induction Hypothesis

Assume that the formula holds true for some positive integer k. That is, assume \(1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + k(k+1) = \frac{k(k+1)(k+2)}{3}\). This is the induction hypothesis.

Step 3: Induction Step

To complete the induction, show that the formula is valid for n = k+1 if it's true for n = k. That is, prove that if \(1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + k(k+1) = \frac{k(k+1)(k+2)}{3}\) then \(1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + (k+1)[(k+1)+1] = \frac{(k+1)[(k+1)+1][(k+1)+2]}{3}\). Substitute the induction hypothesis in order to simplify the left side of the equation and prove the equality.

Key Concepts

Understanding the Base Case

The base case in mathematical induction serves as the foundation for your proof. It verifies that the statement is true for the first value in the sequence, usually when \( n = 1 \). By demonstrating that the equation holds true for this initial value, you establish a starting point for your induction, which can be thought of as the domino effect. In this particular problem, you substitute \( n = 1 \) into both sides of the equation:

• Left Side: \( 1 \times 2 = 2 \)
• Right Side: \( \frac{1(1+1)(1+2)}{3} = 2 \)

Since both sides are equal, the base case holds true. This crucial step shows that the statement works for the first case, supporting the domino principle to carry forward the proof.

Induction Hypothesis Explained

The induction hypothesis assumes the statement is true for some arbitrary positive integer \( k \). This step is like saying, "Let's pretend it's true for \( n = k \), now prove it!" In our case, we assume:

• \( 1 \times 2 + 2 \times 3 + 3 \times 4 + \cdots + k(k+1) = \frac{k(k+1)(k+2)}{3} \)

This assumption is vital, as it allows us to make a connection between one case and the subsequent case. If the assumption holds true for \( k \), then we aim to prove it will also hold for \( k+1 \). This step sets the stage for what comes next in the induction process.

Detailed Induction Step

The induction step is where we prove that if the statement is true for \( n = k \), it must also be true for \( n = k+1 \). This step connects the base case to other positive integers, functioning like the connecting dominoes. We need to demonstrate:

• If \( 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + k(k+1) = \frac{k(k+1)(k+2)}{3} \), then \( 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + (k+1)[(k+1)+1] = \frac{(k+1)[(k+1)+1][(k+1)+2]}{3} \).

To achieve this, add the next term \( (k+1)(k+2) \) to the assumed equation. Replace the left side with the induction hypothesis and simplify:

• Left: \( \frac{k(k+1)(k+2)}{3} + (k+1)(k+2) \)

Factor and simplify to match the right side of the equation. Once both sides are proven equal, the induction step is successful, ensuring that the statement is valid for all positive integers. This process highlights the power and elegance of mathematical induction.