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The expression you provided involves a product and a summation. It is written as:

\[
\prod_{i=1}^{2} \left( \sum_{j=1}^{50} j^i \right)
\]

We can break this down into steps to calculate the value:

1. First, evaluate the summation \(\sum_{j=1}^{50} j^i\) for \(i = 1\) and \(i = 2\).
   - For \(i = 1\), the sum is: \(\sum_{j=1}^{50} j\), which is the sum of the first 50 natural numbers.
   - For \(i = 2\), the sum is: \(\sum_{j=1}^{50} j^2\), which is the sum of the squares of the first 50 natural numbers.

2. Once we calculate the two summations, take the product of both sums.

Let's compute this step by step.The value of the given operation is:

\[
\prod_{i=1}^{2} \left( \sum_{j=1}^{50} j^i \right) = 54,729,375
\]

Breaking it down:
- \(\sum_{j=1}^{50} j = 1275\)
- \(\sum_{j=1}^{50} j^2 = 42,925\)

The product of these two summations is:

\[
1275 \times 42,925 = 54,729,375
\]

Would you like further details or clarifications?

Here are 5 related questions to deepen your understanding:
1. What is the formula for the sum of the first \(n\) natural numbers?
2. How can you generalize the sum of squares of the first \(n\) natural numbers?
3. How would the result change if the range of the summation changed from 50 to another number?
4. What happens if you extend the product beyond \(i = 2\) (i.e., for \(i = 3\))?
5. Can this problem be solved using any mathematical shortcuts or approximations?

**Tip**: The sum of squares formula is \(\sum_{j=1}^n j^2 = \frac{n(n+1)(2n+1)}{6}\). It helps in simplifying calculations for large sums.

Encontrar el valor numérico de la siguiente operación: \prod_{i=1}^{2} \left( \sum_{j=1}^{50} j^i \right)

Learn how to calculate the product of two summations where i ranges from 1 to 2 and j from 1 to 50. This problem covers key mathematical concepts including the sum of natural numbers and the sum of squares of natural numbers. Suitable for advanced high school or early college students.

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