The value of $\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\frac{3}{{{r^2} - r + 9}}} \right)} $ is-

If $\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots \left(1+\frac{2n+1}{n^2}\right) = 121$,then $n =$

The sum $\sum_{n=1}^{21} \frac{3}{(4n-1)(4n+3)}$ is equal to

$\text{Given, } \frac{\sin 1^{\circ}}{\sin x^{\circ} \sin (x+1)^{\circ}} = \cot x^{\circ} - \cot (x+1)^{\circ}, \text{ then the value of } \frac{1}{\sin 45^{\circ} \sin 46^{\circ}} + \frac{1}{\sin 46^{\circ} \sin 47^{\circ}} + \dots + \frac{1}{\sin 89^{\circ} \sin 90^{\circ}} \text{ is}$

For any $n \in N$,$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \ldots + \frac{1}{(3n-1)(3n+2)} = $

If $t_n = \frac{1}{4}(n+2)(n+3)$ for $n = 1, 2, 3, \ldots$,then $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{2003}}$ is equal to