Question

Let $S=\frac{1^2}{1.3}+\frac{2^2}{3.5}+\frac{3^2}{5.7}+......+\frac{{500}^2}{999.1001}$ then the number of positive divisors of $\left[S\right]$ is
(where $[.]$ denotes the greatest integer function)

Answer 1

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

$\sum 4T_r=\sum (\frac{r}{2r-1}+\frac{r}{2r+1})$

$4S=1+(\frac{1}{3}+\frac{2}{3})+(\frac{2}{5}+\frac{3}{5})+---------+\frac{500}{999}+\frac{500}{1001}$

$4S=1+499+\frac{500}{1001}$
[S] = 125
no. of divisors are $\to$ $1,5,25,125$