Math Problem Statement
2021 × 2022 × 2023 × [1 4 × 2 5 × 3 6 × 4 7 × 5 8 × 6 9 × … × 2017 2020 × 2018 2021 × 2019 2022 × 2020 2023 ].
Solution
The expression you've provided is a large product involving two parts. Let's break it down step by step:
- The first part is:
- The second part is a long product, which appears to be a sequence of fractions. This pattern seems to pair numbers in a ratio, from 1/4 to 2020/2023. The second part can be written as:
Now, the key to simplifying this is to recognize that many terms will cancel out.
Simplifying the second part:
The sequence cancels out all the terms in the numerator except the very first and all terms in the denominator except the last few. Let's write this out:
The numerator will be:
The denominator will be:
Therefore, the second part simplifies to:
But, by the properties of factorials:
Final Simplification:
Multiplying the first and second parts together:
Thus, the value of the entire expression is:
Would you like more details or have any questions about this solution?
Here are 5 related questions:
- What are factorials and how are they calculated?
- Can we generalize this type of product for other similar patterns?
- How does cancellation of terms work in a large product of fractions?
- What are some common applications of factorials in mathematics?
- How does the concept of limits apply to factorials?
Tip: When dealing with large products or sequences, always check if terms can cancel each other to simplify the calculation.
Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Factorials
Sequences and Series
Simplification of Fractions
Formulas
n! (Factorial)
Product of Fractions
Theorems
Factorial Simplification Theorem
Cancellation of Terms in Product Sequences
Suitable Grade Level
Grades 11-12 or Undergraduate
Related Recommendation
Evaluating the Product of Fractions from 1/2 to 1/2023
Find the Closest Integer of a Fraction Involving Summation and Product
Closest Integer to \( \frac{2023 \times 2024 \times 2025}{1^2 + 3^2 + \ldots + 2023^2} \)
Solving Recursive Sequence x_n = (n/2) * x_{n-1} and Product of Factorials
Simplifying the Product (1 + 1/2)(1 + 1/3)...(1 + 1/2025) using Telescoping Product