https://file.aiquickdraw.com/mathbot/file/1729756895998tjrjlmkp.jpg

The expression shown in the image is:

\[
(1 - \frac{1}{2^2})(1 - \frac{1}{3^2})(1 - \frac{1}{4^2}) \cdots (1 - \frac{1}{100^2})
\]

This is a product of terms of the form \((1 - \frac{1}{n^2})\) for \(n = 2, 3, \dots, 100\).

Each term can be simplified as:

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

So the full product becomes:

\[
\prod_{n=2}^{100} \frac{(n-1)(n+1)}{n^2}
\]

This expression simplifies through cancellation. Many terms cancel out, leaving:

\[
\frac{1 \cdot 101}{2 \cdot 100} = \frac{101}{200}
\]

Thus, the correct answer is:

\[
\boxed{\frac{101}{200}}
\]

Would you like further details or have any questions? 

Here are 5 related questions for deeper understanding:
1. How does partial fraction decomposition apply to products like this?
2. Can this type of product be generalized for larger values of \( n \)?
3. What other sequences or products exhibit similar cancellation patterns?
4. How does the simplification process change if different powers are used in the denominator?
5. What other types of problems can be solved using telescoping products?

**Tip:** When simplifying products involving fractions, look for patterns of cancellation to reduce the complexity of the expression.

Calculate the expression: (1 - 1/2^2)(1 - 1/3^2)...(1 - 1/100^2)

This problem involves simplifying a telescoping product of fractions of the form (1 - 1/n^2) from n=2 to n=100. Learn how to apply simplification techniques and cancellation patterns to find the correct result of 101/200.

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