(1 squared over 13) + (2squared over 35)+......+50squared over 99*101

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!

(1 squared over 13) + (2 squared over 35) + ... + 50 squared over 99*101

Explore how to solve a summation problem involving the terms (n^2 / (2n-1)(2n+1)) where n ranges from 1 to 50. This detailed breakdown uses algebraic techniques to simplify the sum and provide a numerical approximation for the result. Suitable for high school students in Grades 10-12.

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