Problem 18 - A Graphical Approach to Precalculus with Limits

Question

Use mathematical induction to prove each statement. Assume that \(n\) is a positive integer. $$1 \cdot 2+2 \cdot 3+\dots+n(n+1)=\frac{n(n+1)(n+2)}{3}$$

Step-by-Step Solution

Answer

The statement is proven using mathematical induction.

Step 1: Understand the Problem

We need to prove the formula \(1 \cdot 2 + 2 \cdot 3 + \dots + n(n+1) = \frac{n(n+1)(n+2)}{3}\) using mathematical induction. The process involves showing it holds for the base case and assuming it holds for an arbitrary case \(k\), then proving it for \(k+1\).

Step 2: Prove Base Case (n=1)

For \(n=1\), the left side of the equation becomes \(1 \cdot 2 = 2\). The right side becomes \(\frac{1(1+1)(1+2)}{3} = \frac{2 \times 3}{3} = 2\). Thus, the base case holds true.

Step 3: Inductive Hypothesis

Assume the formula is true for \(n = k\), i.e., \(1 \cdot 2 + 2 \cdot 3 + \dots + k(k+1) = \frac{k(k+1)(k+2)}{3}\). This is our inductive hypothesis.

Step 4: Prove for n=k+1

We need to show that the formula holds for \(n = k+1\). The left side becomes \((1 \cdot 2 + 2 \cdot 3 + \dots + k(k+1)) + (k+1)(k+2)\). From the inductive hypothesis, substitute: \(\frac{k(k+1)(k+2)}{3} + (k+1)(k+2)\).

Step 5: Simplify the Expression

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

Step 6: Conclusion

After simplification, the formula holds for \(k+1\):\[ 1 \cdot 2 + 2 \cdot 3 + \cdots + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3} \]By mathematical induction, the original statement is proved.

Key Concepts

Proofs in Mathematics

Proofs are fundamental to understanding and establishing the truth of mathematical statements. In mathematics, a proof demonstrates why a particular statement or formula is universally true, not just a random occurrence. When you embark on proving something, you are following a logical sequence of statements to derive a conclusion from initial assumptions or previously proven results.

Some important aspects of proofs in mathematics include:

• Clarity and precision: A proof must be clear and precise so that others can follow your reasoning.
• Logical flow: Each step in the proof should follow logically from the previous steps.
• Use of established facts: Proofs often rely on known results and axioms as building blocks.

Understanding proofs helps you think critically and approach mathematical problems in a structured manner. It's like solving a puzzle where each piece is vital for the bigger picture. And different methods, like direct proof, indirect proof, and proof by induction, give us tools adapted to specific types of problems.

Inductive Hypothesis

The inductive hypothesis is a key component in the method of mathematical induction. When using induction, you assume that a given statement is true for an arbitrary positive integer, typically denoted by \(k\). This assumption is called the inductive hypothesis. You then show that if the statement holds for \(k\), it must also be true for the next integer, \(k+1\).

The steps are as follows:

• Assumption for \(n = k\): Assume the original statement is true for \(n = k\). For instance, in our problem, we assume that \(1 \cdot 2 + 2 \cdot 3 + \dots + k(k+1) = \frac{k(k+1)(k+2)}{3}\).
• Proof for \(n = k+1\): Using the assumption for \(n = k\), prove that the statement holds for \(n = k+1\). You adapt the expression to include the \((k+1)\)th term and simplify using your assumption.

The logic behind this is that if your assumption leads inevitably to the next step being true, you are effectively creating a domino effect that proves the statement for all positive integers, starting from an initial step. This approach gives us a powerful tool to tackle problems involving sequences or series.

Base Case in Induction

The base case in induction is the foundation upon which the rest of the inductive proof is built. It is the first step where you check that the statement is true for the initial value of the sequence or series, usually with \(n = 1\).

The importance of the base case can be summarized as follows:

• Starting point: It proves that the statement holds true at the beginning of the sequence.
• Verification: By confirming the base case, you establish that there is at least one starting point where the statement is true.
• Enables induction: With the base case proven, you set the stage for using the inductive hypothesis to prove the statement for all subsequent cases.

In our example, the base case was checking the statement for \(n = 1\), where both sides of the equation equaled 2, confirming the formula's correctness at \(n = 1\). Without this step, the entire induction process could fall apart. It's like proving the first domino in a chain will fall, ensuring the remaining dominoes can follow suit.