To solve this problem, let's start by understanding the recursive formula provided and the expression $2024 \prod_{i=1}^n i$.

Understanding the Recursive Formula

Given:

• $x_1 = 2024$
• $x_n = \frac{n}{2} x_{n-1}$ for $n > 1$

This recursive relation shows that each term $x_n$ is calculated based on the previous term $x_{n-1}$, multiplied by $\frac{n}{2}$.

Calculating the General Expression for $x_n$

Let's compute the first few values to identify a pattern:

x_1 &= 2024, \\ x_2 &= \frac{2}{2} x_1 = 1 \times 2024 = 2024, \\ x_3 &= \frac{3}{2} x_2 = \frac{3}{2} \times 2024 = 3036, \\ x_4 &= \frac{4}{2} x_3 = 2 \times 3036 = 6072, \\ x_5 &= \frac{5}{2} x_4 = \frac{5}{2} \times 6072 = 15180. \end{align*}$$ We see that the formula indeed builds upon the previous term multiplied by $$\frac{n}{2}$$. ### Expression $$2024 \prod_{i=1}^n i$$ The expression $$2024 \prod_{i=1}^n i$$ involves multiplying 2024 by the product of all integers from 1 to $$n$$. This can be written as: $$2024 \times (1 \times 2 \times 3 \times \ldots \times n) = 2024 \times n!.$$ This shows that the expression is 2024 multiplied by the factorial of $$n$$. ### Conclusion The result for $$x_n$$ in terms of the recursive formula involves repeatedly applying the multiplication by $$\frac{n}{2}$$, and the expression $$2024 \prod_{i=1}^n i$$ is equal to $$2024 \times n!$$. ### Follow-up Questions: 1. Do you want to find the value of $$x_n$$ for a specific $$n$$? 2. Are you interested in the derivation of a closed formula for $$x_n$$? 3. Would you like to explore how factorial growth impacts the values of $$x_n$$? 4. Do you want to understand the convergence properties of this sequence? 5. Would you like to see the computation of $$x_n$$ for higher values of $$n$$? #### Tip: When dealing with recursive sequences, try calculating the first few terms to identify patterns that can help generalize the formula!