The sum of the series $\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots$ up to $n$ terms is equal to:

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 $.....$

$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \ldots + \frac{1}{(3n-1)(3n+2)}$ is equal to

$1 + \sum\limits_{r = 0}^{22} {\left\{ {r\left( {r + 2} \right) + 1} \right\}} \cdot r! = k!$,then the number of divisors of $k$ is

Find the sum to $n$ terms of the series: $\frac{3}{1 \cdot 2} \cdot \frac{1}{2} + \frac{4}{2 \cdot 3} \cdot \left( \frac{1}{2} \right)^2 + \frac{5}{3 \cdot 4} \cdot \left( \frac{1}{2} \right)^3 + \dots$

If ${t_n} = \frac{1}{4}(n + 2)(n + 3)$ for $n = 1, 2, 3, \dots$,then $\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \dots + \frac{1}{t_{2003}} = $