https://file.aiquickdraw.com/mathbot/file/1726395791736ejn4eoqb.jpg

The problem in the image shows a product of expressions of the form:

\[
(1 + \frac{1}{2})(1 + \frac{1}{3})(1 + \frac{1}{4}) \dots (1 + \frac{1}{98})(1 + \frac{1}{99})
\]

This can be rewritten as:

\[
\prod_{n=2}^{99} \left( 1 + \frac{1}{n} \right)
\]

Each term \( 1 + \frac{1}{n} \) can be written as:

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

So, the whole product becomes:

\[
\prod_{n=2}^{99} \frac{n+1}{n} = \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \dots \times \frac{100}{99}
\]

In this telescoping product, many terms cancel out, leaving only:

\[
\frac{100}{2} = 50
\]

Thus, the value of the product is \( \boxed{50} \).

Would you like more details or have any questions?

Here are 5 related questions for practice:

1. What is the general form of a telescoping product?
2. How does the telescoping effect simplify large products?
3. What happens if the product starts from a different index, say \( n=1 \)?
4. How would you handle a similar problem with more complex terms?
5. How does the final result change if one more factor is added to the product?

**Tip:** In telescoping products, always look for cancellations that simplify the expression!

How much is (1 + 1/2)(1 + 1/3)(1 + 1/4)...(1 + 1/98)(1 + 1/99)?

Discover how to solve a telescoping product of fractions, from (1 + 1/2) to (1 + 1/99), with step-by-step simplification and cancellation techniques. This problem is ideal for students in Grades 10-12, covering key algebraic concepts and formulas.

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