To solve the problem presented in the image, we need to find the closest integer for the given fraction:

$\frac{2023 \cdot 2024 \cdot 2025}{1^2 + 3^2 + 5^2 + \dots + 2021^2 + 2023^2}.$

Here is the step-by-step approach:

Step 1: Numerator Calculation

The numerator is the product: $2023 \cdot 2024 \cdot 2025.$

This can be expanded using properties of consecutive integers: $2023 \cdot 2024 \cdot 2025 = 2024 \cdot (2023 \cdot 2025).$ Simplify $2023 \cdot 2025$ as: $2023 \cdot 2025 = (2024 - 1)(2024 + 1) = 2024^2 - 1.$ Thus, the numerator becomes: $2024 \cdot (2024^2 - 1) = 2024^3 - 2024.$

Step 2: Denominator Calculation

The denominator is the sum of squares of all odd numbers up to $2023$: $1^2 + 3^2 + 5^2 + \dots + 2023^2.$

The sum of squares of the first $n$ odd numbers is given by: $1^2 + 3^2 + 5^2 + \dots + (2n-1)^2 = \frac{n(2n-1)(2n+1)}{3}.$

Here, the largest odd number is $2023$, so: $2023 = 2n - 1 \implies n = 1012.$

Substitute $n = 1012$ into the formula: $1^2 + 3^2 + \dots + 2023^2 = \frac{1012 \cdot 2023 \cdot 2025}{3}.$

Step 3: Combine the Results

The fraction simplifies to: $\frac{2023 \cdot 2024 \cdot 2025}{\frac{1012 \cdot 2023 \cdot 2025}{3}}.$

Cancel out common terms $2023$ and $2025$: $= \frac{2024 \cdot 3}{1012}.$

Simplify further: $= \frac{2024}{1012} \cdot 3 = 2 \cdot 3 = 6.$

Final Answer:

The closest integer is: $\boxed{6}.$

Would you like me to provide further clarification or details?

5 Related Questions:

1. What is the formula for the sum of squares of the first $n$ integers?
2. How do you derive the formula for the sum of squares of odd numbers?
3. Can this type of problem be solved using approximations or inequalities?
4. How do products of consecutive integers simplify algebraically?
5. What are some practical applications of such summation problems?

Tip:

For problems involving sums of sequences, always check if a known formula applies—it can save a lot of computation time!