The problem shown in the image involves a product of terms of the form:

$\left( 1 + \frac{4}{3} \right)\left( 1 + \frac{4}{7} \right)\left( 1 + \frac{4}{11} \right) \dots \left( 1 + \frac{4}{95} \right)$

Let's express each term in the product. Each term takes the form:

$1 + \frac{4}{n} = \frac{n + 4}{n}$

Thus, the product becomes:

$\frac{3 + 4}{3} \times \frac{7 + 4}{7} \times \frac{11 + 4}{11} \times \dots \times \frac{95 + 4}{95} = \frac{7}{3} \times \frac{11}{7} \times \frac{15}{11} \times \dots \times \frac{99}{95}$

Notice that this is a telescoping product, meaning most terms will cancel out. Let's simplify:

$\frac{7}{3} \times \frac{11}{7} \times \frac{15}{11} \times \dots \times \frac{99}{95}$

All intermediate terms cancel, and we are left with:

$\frac{99}{3} = 33$

Thus, the answer is $\boxed{33}$.

Would you like more details on how telescoping products work or have any other questions?

Here are 5 related questions:

1. What are telescoping sequences, and how are they used to simplify products or sums?
2. How can this product be generalized to other types of series or fractions?
3. What happens when a similar product has terms with alternating signs?
4. Can you derive a general formula for such telescoping products with other coefficients?
5. How would this problem change if there were additional terms in the product?

Tip: Always look for patterns like telescoping when dealing with long products or sums, as they can simplify complex expressions significantly.