91Ó°ÊÓ

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.

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: \[C_{n-1}=3 C_{n-2}+\left(C_{n-2}-\frac{6}{n} C_{n-2}\right)\]

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 \(n \geq 1\).