b=(1/2+1)(1/3+1)(1/4+1)................(1/99+1)

You are tasked with evaluating the product:

\[
b = \left( \frac{1}{2} + 1 \right) \left( \frac{1}{3} + 1 \right) \left( \frac{1}{4} + 1 \right) \dots \left( \frac{1}{99} + 1 \right)
\]

Let's first simplify each term in the product. The general term for \( n \) is:

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

Thus, the full product becomes:

\[
b = \prod_{n=2}^{99} \frac{n+1}{n}
\]

We can now break down the product:

\[
b = \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \dots \times \frac{100}{99}
\]

Notice that this is a **telescoping product**, where most of the terms cancel out. Specifically, all intermediate terms cancel each other, leaving:

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

Thus, the value of \( b \) is:

\[
b = 50
\]

Would you like to see more steps or have any further questions on this?

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Here are some related questions you may find useful:
1. What is the principle behind telescoping series/products?
2. How do you evaluate a product like this when there are many terms?
3. What happens when the terms don't cancel out completely in a telescoping product?
4. How would you approach a similar product but with different terms, such as \( \prod_{n=2}^{99} \frac{n+2}{n} \)?
5. What are some applications of telescoping sums and products in mathematics?

**Tip**: In telescoping products, always look for patterns where terms can cancel each other out, making the calculation much simpler!

(1/2 + 1)(1/3 + 1)(1/4 + 1) ... (1/99 + 1)

Learn how to evaluate the telescoping product (1/2 + 1)(1/3 + 1)(1/4 + 1) ... (1/99 + 1) and simplify it into a much simpler form. This step-by-step solution explains the telescoping product concept and how most terms cancel out to give the final result of 50. Suitable for high school students.

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