https://file.aiquickdraw.com/mathbot/file/1725706825843nayfcxcw.jpg

The expression presented in the image is:

\[
\left( 1 - \frac{1}{5} \right) \left( 1 - \frac{1}{6} \right) \left( 1 - \frac{1}{7} \right) \cdots \left( 1 - \frac{1}{2009} \right)
\]

Let's simplify the general term of this product:

\[
1 - \frac{1}{n} = \frac{n - 1}{n}
\]

Thus, the whole expression becomes:

\[
\frac{4}{5} \times \frac{5}{6} \times \frac{6}{7} \times \cdots \times \frac{2008}{2009}
\]

Notice that the terms cancel out in a telescoping pattern:

\[
\frac{4}{5} \times \frac{5}{6} \times \frac{6}{7} \times \cdots \times \frac{2008}{2009} = \frac{4}{2009}
\]

So, the value of the expression is:

\[
\frac{4}{2009}
\]

Looking at the answer choices provided:

A) \(\frac{4}{2009}\)

B) \(\frac{5}{2009}\)

C) \(\frac{2}{2009}\)

D) \(\frac{2}{2000}\)

The correct answer is **A) \(\frac{4}{2009}\)**.

Would you like more details on this solution or any other clarifications?

Here are 5 related questions for practice:
1. How do you simplify products of similar fractional terms?
2. What is the telescoping property in sequences?
3. How would the expression change if it started from \(n = 4\) instead of \(n = 5\)?
4. What if the sequence ended at 2010? How would that affect the result?
5. How would you compute this product if the denominators were squared?

**Tip**: Telescoping sequences are very useful in simplifying large products or sums where intermediate terms cancel out.

Evaluate the product (1 - 1/5)(1 - 1/6)(1 - 1/7) ... (1 - 1/2009)

Learn how to simplify a telescoping product involving fractions from 1 - 1/5 to 1 - 1/2009. This problem utilizes key concepts in algebra and demonstrates how terms cancel out in a telescoping sequence.

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