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)

Open in App

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 $\to$ $1, 5, 25, 125$

Suggest Corrections

Similar questions

S = \frac{1^{2}}{1.3} + \frac{2^{2}}{3.5} + \frac{3^{2}}{5.7} + . . . . . . + \frac{500^{2}}{999.1001}

[S]

[.]

S = \sum_{k = 1}^{80} \frac{1}{\sqrt{K} .}

sin x = [x]

[\cdot]

S = (1 + 0.0001)^{1000}

[S]

sin x = [x]

.

Join BYJU'S Learning Program