$\frac{1^2}{1\cdot3} + \frac{2^2}{3\cdot5} + \frac{3^2}{5\cdot7}+\cdots+\frac{500^2}{999\cdot1001} = ?$ I found this problem in a high school text book.

Let $ \displaystyle s = \frac{1^2}{1\cdot3} + \frac{2^2}{3\cdot5} + \frac{3^2}{5\cdot7}+\cdots+\frac{500^2}{999\cdot1001}$. Find $s$.

How I tried:
Observe that  $T_n = \frac{n^2}{(2n-1)(2n+1)}$. Here, $T_n$ is the $n$th  term of the sequence. So, we  need to find the value of $\sum_{n=1}^{500}\frac{n^2}{(2n -1)(2n+1)}$. We can find its value with Telescope Cancellation Method, but it requires breaking the  expression we got into simpler terms.
How to simplify  $T_n = \frac{n^2}{(2n-1)(2n+1)}$ ?
 A: $$T_n = \frac{n^2}{4n^2-1} = \frac{1}{4}\left( \frac{4n^2-1+1}{4n^2-1} \right) = \frac{1}{4}\left( 1 + \frac{1}{4n^2-1} \right)=\frac{1}{4}\left( 1 + \frac{1}{(2n-1)(2n+1)} \right). $$
Decomposing with partial fractions, we have:
$$ \frac{1}{(2n-1)(2n+1)} = \frac{\frac{1}{2}}{2n-1} - \frac{\frac{1}{2}}{2n+1} = \frac{1}{2}\left( \frac{1}{2n-1} - \frac{1}{2n+1} \right). $$
And we see that most of the terms in $\displaystyle\sum_{n=1}^{500} T_n$ will cancel, apart from a few at either end.
A: I don't think your $T_n$ is correct. Here's how I broke it down:
$$\frac{1^2}{1.3} + \frac{2^2}{3.5} + \frac{3^2}{5.7}+ ... = \frac{1^2}{1 + 0.1 \cdot 3} + \frac{2^2}{3 +0.1 \cdot 5} + \frac{3^2}{5+ 0.1 \cdot 7} +  ... 
\\ = \sum_{n=1}^N \frac{n^2}{(2n-1)+0.1(2n+1)}
\\ = \sum_{n=1}^N \frac{10n^2}{10(2n-1)+(2n+1)}
\\ = \sum_{n=1}^N \frac{10n^2}{22n-9}
$$
From here, it becomes a matter of polynomial long division, which yields:
$$\sum_{n=1}^N \left(\frac{5n}{11} + \frac{45}{242} +\frac{\frac{405}{242}}{22n-9} \right)
\\= \frac{5N(N+1)}{22}+\frac{45N}{242}+\frac{405}{242} \cdot \sum_{n=1}^N \frac{1}{22n-9}
$$
Not the prettiest expression, I know. What's more, there exists no closed-form expression for that final sum. But if you plug this in for $N = 500$ into a calculator, it will yield $\approx 57025.3858937$
