The question asks to calculate the product of a series of terms:

$\left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right)\left(1 + \frac{1}{4}\right) \cdots \left(1 + \frac{1}{98}\right)\left(1 + \frac{1}{99}\right)$

Each term in the product follows the form $1 + \frac{1}{n}$, which simplifies to $\frac{n+1}{n}$. The entire expression becomes:

$\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \cdots \times \frac{99}{98} \times \frac{100}{99}$

Notice that this is a telescoping product, meaning that most of the terms will cancel out. Specifically, all terms in the numerator and denominator cancel except for the very first denominator and the very last numerator. Therefore, the entire product simplifies to:

$\frac{100}{2} = 50$

Thus, the value of the expression is $50$.

Do you need further details, or have any questions?

Here are 5 related questions:

1. What happens if the series continues beyond 99 terms?
2. How does the concept of a telescoping product apply to sums?
3. Can this series be represented using factorials?
4. What is the general solution to a telescoping product?
5. How does this problem change if the terms involve subtraction instead of addition?

Tip: Always look for cancellation opportunities in large products to simplify calculations.