If $a_1, a_2, a_3, ..., a_n$ is an arithmetic progression,then $\frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + \frac{1}{a_3 a_4} + ... + \frac{1}{a_{n-1} a_n} = ...$

If $t_{n} = \frac{1}{4}(n+2)(n+3)$,$n \in N$,then which one of the following is true?
Assertion $(A)$ : $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{2003}} = \frac{2003}{3009}$
Reason $(R)$ : $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{n}} = \frac{4n}{3(n+3)}$

The sum of $10$ terms of the series $\frac{3}{1^{2} \times 2^{2}}+\frac{5}{2^{2} \times 3^{2}}+\frac{7}{3^{2} \times 4^{2}}+\ldots$ is :

The value of $\frac{1}{(1 + a)(2 + a)} + \frac{1}{(2 + a)(3 + a)} + \frac{1}{(3 + a)(4 + a)} + \dots + \infty$ is,(where $a$ is a constant)

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 $.....$

Find the sum of the series $1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + \dots$ up to $n$ terms.