https://file.aiquickdraw.com/mathbot/file/17295518835016ctbz0q7.jpg

The expression in the image you provided is the following product:

\[
\left(1 - \frac{1}{2^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{4^2}\right)\dots \left(1 - \frac{1}{49^2}\right)\left(1 - \frac{1}{50^2}\right)
\]

This can be simplified step-by-step:

Each factor follows the pattern:

\[
\left(1 - \frac{1}{n^2}\right) = \frac{n^2 - 1}{n^2} = \frac{(n-1)(n+1)}{n^2}
\]

Therefore, the entire product is:

\[
\prod_{n=2}^{50} \left( \frac{n-1}{n} \cdot \frac{n+1}{n} \right)
\]

This expression forms a telescoping product, meaning that most of the terms will cancel out. Let's explore how:

- For \( n = 2 \), the first term is \( \frac{1 \cdot 3}{4} \),
- For \( n = 3 \), the second term is \( \frac{2 \cdot 4}{9} \), and so on.

Eventually, almost all terms cancel out, except the first and last terms of the sequence.

The simplified expression is:

\[
\frac{1 \cdot 51}{2 \cdot 50}
\]

So, the final result of the product is:

\[
\frac{51}{100} = 0.51
\]

Thus, the value of the given product is \( 0.51 \).

Would you like more details, or do you have any questions?

Here are 5 related questions for further exploration:
1. How do you recognize a telescoping series or product?
2. What is the general form of a product involving terms like \(1 - \frac{1}{n^2}\)?
3. How can you find the sum of a telescoping series instead of a product?
4. How do these types of expressions arise in number theory or other areas of mathematics?
5. How can you generalize this for an arbitrary upper bound (not just 50)?

**Tip**: When dealing with products of fractions or sums, look for patterns such as telescoping to simplify the expression.

Simplify the product: (1 - 1/2^2)(1 - 1/3^2)(1 - 1/4^2)...(1 - 1/49^2)(1 - 1/50^2).

This guide simplifies the product (1 - 1/2^2)(1 - 1/3^2)...(1 - 1/50^2) by recognizing it as a telescoping series. Learn how to systematically cancel intermediate terms to reach the final simplified value of 0.51.

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