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 $[S]$ is
(where $[.]$ denotes the greatest integer function)

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Solution

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

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

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

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

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