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.

What is the sum of $n$ terms of the series $1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + \dots$?

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

If ${a_1}, {a_2}, \dots, {a_{n+1}}$ are in $A.P.$,then $\frac{1}{{{a_1}{a_2}}} + \frac{1}{{{a_2}{a_3}}} + \dots + \frac{1}{{{a_n}{a_{n+1}}}}$ is

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 $(R)$ : $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{n}} = \frac{4n}{3(n+3)}$

If $\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots+\frac{1}{(180-a)(200-a)}=\frac{1}{256}$,then the maximum value of $a$ is.