$\frac{1^2}{1\cdot3} + \frac{2^2}{3\cdot5} + \frac{3^2}{5\cdot7}+\cdots+\frac{500^2}{999\cdot1001} = ?$ 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  $T_n = \frac{n^2}{(2n-1)(2n+1)}$ ?

$$T_n = \frac{n^2}{4n^2-1} = \frac{1}{4}\left( \frac{4n^2-1+1}{4n^2-1} \right) = \frac{1}{4}\left( 1 + \frac{1}{4n^2-1} \right)=\frac{1}{4}\left( 1 + \frac{1}{(2n-1)(2n+1)} \right). $$
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.
