Question

Prove the following statements with either induction, strong induction or proof by smallest counterexample. If \(n \in \mathbb{N},\) then \(1 \cdot 3+2 \cdot 4+3 \cdot 5+4 \cdot 6+\cdots+n(n+2)=\frac{n(n+1)(2 n+7)}{6}\).

Short answer

The proof can be completed using the method of mathematical induction. None of the steps require a huge jump in logic, but the inductive step can be slightly tricky as it involves precise algebraic manipulations. The proof relies heavily on building upon the base case, and using that in the induction hypothesis to finally prove for the n = k + 1 case. With the completion of these steps, the statement is successfully proven.

Step-by-step solution

Base Case Verification

Start by validating the expression for n = 1. On the left side of the equation we have: \(1(3)\) which equals 3. On the right side of the equation we get: \(\frac{1(1+1)(2(1)+7)}{6}\) which also simplifies to 3.

Induction Hypothesis

Assume the given statement holds true for an arbitrary positive integer n = k, i.e., \(1(3) + 2(4) +...+ k(k+2) = \frac{k(k+1)(2k+7)}{6}\). This is our assumption and we will use this in the next step.

Inductive Step

The goal here is to prove the statement for n = k + 1, i.e. \(1(3) + 2(4) +...+ k(k+2) + (k+1)((k+1)+2) = \frac{(k+1)(k+1+1)(2(k+1)+7)}{6}\). Add \((k+1)((k+1)+2)\) to both sides of our hypothesized equation. On the left side, this gives the full left hand side of our target equation. On the right side, perform the necessary algebraic manipulations to derive the right hand side of our target equation.

Key concepts

Induction Proof

Understanding an induction proof is akin to watching a row of dominoes topple one after another. Once you're certain the first one falls (the base case), and you've ascertained that every single domino can knock over the next one (the inductive step), you can be confident that all the dominoes will fall.

In mathematical terms, induction allows us to prove statements that are typically formulated 'for every natural number n'. Just as we checked the base case for n=1 in our exercise and found that it held true, you'll similarly begin by ensuring that the initial condition meets the statement. Then, you have your induction hypothesis: this is where you suppose that the statement works for an arbitrary natural number k. You're not claiming it's true; think of it as a 'what if' scenario to help you move forward.

From there, you jump into the heart of the induction by showing that if the statement holds for k, it must also be true for k+1. If you can demonstrate this step successfully, the proof stands. Throughout the process, algebraic manipulations play a crucial role in verifying each step, much like the solution method we employed in our exercise.

Strong Induction

Imagine, for a moment, that instead of guaranteeing each domino knocks only the next in line, you ensure that one domino can topple all following pieces up to a certain point. That's the essence of strong induction, a robust variation of the typical induction proof.

Strong induction differs slightly in that the induction hypothesis assumes the statement to be true for all integers less than or equal to an arbitrary positive integer k, rather than for just k. The inductive step then demonstrates that the statement is true for k+1 based on this extended hypothesis. This method often comes into play when the problem requires establishing the validity of a statement reliant on several preceding cases, not just the immediate predecessor.

Proof by Smallest Counterexample

The proof by smallest counterexample is a stylish, defensive play in the mathematician's playbook. Suppose you want to assert that a statement is true, but someone challenges you by suggesting there's an exception. To refute them, you would demonstrate that if there were a smallest counterexample, it would lead to a contradiction, thus proving that no counterexample can exist.

Typically, you would assume that there is a least natural number for which the statement fails. By using logic and properties of the integers, you work to derive a contradiction or to show that there must be an even smaller counterexample, which conflicts with our original 'smallest' assumption. This kind of proof is especially useful when direct induction is impractical or when the statement's validity is more convoluted.

Mathematical Proofs

Mathematical proofs are the foundation on which the edifice of mathematics is built. They provide the rigorous justification for why things are true in a logical and systematic way. Proofs can come in many flavors: some are direct, some are indirect (like proof by contradiction), and others use specific techniques like induction, as we have seen.

The journey of proving a mathematical statement often begins with an exploration of definitions and known theorems. By connecting these dots through logic and creative thinking, a mathematician arrives at new truths. The proof, in its written form, communicates this journey in a clear, step-by-step fashion, allowing others to verify and understand the reasoning behind the conclusion. As we showed in the workbook exercise, a proof by induction anchors itself by establishing a base case and an inductive step, forming an airtight argument that leaves no room for doubt regarding the statement's veracity across all natural numbers.