The expression you've provided involves summing terms of the form $\frac{n^2}{(2n-1)(2n+1)}$, where $n$ ranges from 1 to 50. Let's break this down:

The general term in the sum can be rewritten as:

$\frac{n^2}{(2n-1)(2n+1)} = \frac{n^2}{4n^2 - 1}$

This means your full sum is:

$S = \sum_{n=1}^{50} \frac{n^2}{(2n-1)(2n+1)} = \sum_{n=1}^{50} \frac{n^2}{4n^2 - 1}$

I'll calculate this sum numerically for you.The value of the sum:

$S = \sum_{n=1}^{50} \frac{n^2}{(2n-1)(2n+1)} \approx 12.624$

Let me know if you'd like a breakdown of the method or have further questions. Here are some related problems to consider:

1. How does the sum change if you extend it to $n = 100$?
2. What if you simplify each term differently, such as $\frac{1}{(2n-1)(2n+1)}$?
3. Can you find a closed-form expression for this sum?
4. What happens if the terms were cubed instead of squared?
5. How would you approach this sum if you were given continuous variables?

Tip: Recognizing patterns in sums involving fractions can often lead to simplifications or telescoping series!