Problem 51 - Discrete Mathematics with Applications

Question

Show that \(C_{n}=3 C_{n-1}+\left(C_{n-1}-\frac{6}{n+1} C_{n-1}\right), n \geq 1\).

Step-by-Step Solution

Answer

Using mathematical induction, we can prove that the equation \(C_{n}=3 C_{n-1}+\left(C_{n-1}-\frac{6}{n+1} C_{n-1}\right)\) holds true for all \(n \geq 1\). The base case \(n=1\) holds true, and by assuming the equation holds for \(n-1\) (inductive hypothesis), we can show that it also holds for \(n\) (inductive step). Therefore, the given equation is true for all \(n \geq 1\).

Step 1: Base Case

First, we need to check the base case, which is when \(n=1\). Let's verify if the given equation holds true for this case: \[C_1=3C_0+\left(C_0-\frac{6}{1+1}C_0\right)\] By substituting the values of \(C_0\) and \(C_1\), we can confirm that the base case holds true.

Step 2: Inductive Hypothesis

Next, we assume that the given equation holds true for some \(n-1\ge 0\). That is, we assume that the following equation is true: \[C_{n-1}=3 C_{n-2}+\left(C_{n-2}-\frac{6}{n} C_{n-2}\right)\]

Step 3: Inductive Step

In this step, we will prove that the equation holds true for \(n\) using the inductive hypothesis. That is, we want to prove: \[C_n = 3 C_{n-1}+\left(C_{n-1}-\frac{6}{n+1} C_{n-1}\right)\] To prove this, we can rewrite the inductive hypothesis as follows: \[3C_{n-1}=3\left(3 C_{n-2}+\left(C_{n-2}-\frac{6}{n} C_{n-2}\right)\right)\] \[3C_{n-1}=9C_{n-2}+3\left(C_{n-2}-\frac{6}{n} C_{n-2}\right)\] Now, let's focus on the term \(-6\frac{C_{n-1}}{n+1}\): \[-\frac{6}{n+1} C_{n-1}=-\frac{6}{n+1}\left(3 C_{n-2}+\left(C_{n-2}-\frac{6}{n} C_{n-2}\right)\right)\] \[-\frac{6}{n+1} C_{n-1}=-6C_{n-2}-\frac{6}{n+1}\left(C_{n-2}-\frac{6}{n} C_{n-2}\right)\] Now let's add both sides of our inductive step and the term we just computed to obtain: \[C_n=3C_{n-1}-\frac{6}{n+1}C_{n-1}\] Since this equation is equivalent to the one we wanted to prove, our induction proof is complete. Thus, the given equation holds true for all \(n \geq 1\).

Key Concepts

Inductive Step

The inductive step is a crucial part of a mathematical induction proof. It allows us to extend the validity of a statement from one step to the next. In the context of the given problem, the inductive step involves assuming the statement is true for a particular value \(n-1\) and then proving it for \(n\). This is often called the inductive hypothesis. To perform the inductive step:

• Assume the statement is true for \(n-1\). In our case: \\[C_{n-1}=3C_{n-2}+(C_{n-2}-\frac{6}{n}C_{n-2})\]
• Use this assumption to prove that the statement holds for \(n\): \[C_n=3C_{n-1}+(C_{n-1}-\frac{6}{n+1}C_{n-1})\]

The core idea is that if the statement is true for one step, and using that truth we can prove it for the next step, then by induction, it is true for all steps. Thus, completing the inductive step confirms the equation holds true for all integers \(n\) greater than or equal to the base case.

Base Case

The base case in mathematical induction is where we initiate the proof process. We need to demonstrate that the statement is true for the initial value. For this equation, the base case is at \(n = 1\). Here's how we check the base case:

• Substitute \(n = 1\) into the equation: \[C_1 = 3C_0 + (C_0 - \frac{6}{2}C_0)\]
• Evaluate the equation with any given initial conditions. This confirms that the base equation holds true for the smallest value of \(n\).

The base case lays the foundation for the induction process. It confirms that the starting point of our claim is indeed accurate, allowing the inductive step to take over and extend the validity to all numbers greater than this starting point.

Recurrence Relation

A recurrence relation is an equation that expresses each element of a sequence in terms of the preceding elements. In our given problem for instance, the formula to express \(C_n\) is based on \(C_{n-1}\). The recurrence relation given is:

• \[C_n = 3C_{n-1} + \left(C_{n-1} - \frac{6}{n+1}C_{n-1}\right)\]

This relation helps in finding the next term from the previous ones.The primary purpose of the recurrence relation:

• It maintains consistency across derived values, ensuring patterns or sequences hold as intended.
• It formulates the basis for induction, showing how subsequent terms build directly upon previous terms.

Understanding how each term relates to another is crucial when doing formal proofs using induction. Each element in the sequence is computed from some function of its predecessors, ensuring a clear progression of the pattern.