JEE maths#18

$\begin{aligned} S & = \frac {1^2}{1 \times 3} + \frac {2^2}{3 \times 5} + \frac {3^2}{5 \times 7} + \cdots + \frac {500^2}{999 \times 1001} \\ & = \sum_{n=1}^{500} \frac {n^2}{(2n-1)(2n+1)} \\ & = \sum_{n=1}^{500} \frac {n^2}{4n^2-1} \\ & = \frac 14 \sum_{n=1}^{500} \frac {4n^2-1+1}{4n^2-1} \\ & = \frac 14 \sum_{n=1}^{500} \left(1+ \frac 1{4n^2-1}\right) \\ & = \frac {500}4 + \frac 14 \sum_{n=1}^{500} \frac 1{(2n-1)(2n+1)} \\ & = 125 + \frac 14 \sum_{n=1}^{500} \frac 1{(2n-1)(2n+1)} \\ & = 125 + \frac 18 \sum_{n=1}^{500} \left(\frac 1{2n-1} - \frac 1{2n+1} \right) \\ & = 125 + \frac 18 \left(\frac 11 - \frac 1{1001} \right) \end{aligned}$

$\implies \lfloor S \rfloor = 125 = 5^{\color{#3D99F6}3}$ . Therefore, the number of divisors of $\lfloor S \rfloor$ is ${\color{#3D99F6}3}+1=\boxed{4}$ .