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

Let for $n = 1, 2, \ldots, 50$,$S_{n}$ be the sum of the infinite geometric progression whose first term is $n^{2}$ and whose common ratio is $\frac{1}{(n+1)^{2}}$. Then the value of $\frac{1}{26} + \sum_{n=1}^{50} \left(S_{n} + \frac{2}{n+1} - n - 1\right)$ is equal to

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

If $\frac{1}{2 \times 3 \times 4} + \frac{1}{3 \times 4 \times 5} + \frac{1}{4 \times 5 \times 6} + \dots + \frac{1}{100 \times 101 \times 102} = \frac{k}{101}$,then $34k$ is equal to $.....$

$\frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \ldots$ to $24$ terms $=$

If $a_1, a_2, \dots, a_n$ are in $A.P.$ with common difference $d$,then the sum of the series $\sin d (\csc a_1 \csc a_2 + \csc a_2 \csc a_3 + \dots + \csc a_{n-1} \csc a_n)$ is: