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

Statement-$1$: The sum of the series $1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)$ is $8000$.
Statement-$2$: For any natural number $n$,$\sum_{k=1}^n (k^3 - (k-1)^3) = n^3$.

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.

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

The sum of $(n + 1)$ terms of $\frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots$ is

If $\frac{1}{2 \times 4} + \frac{1}{4 \times 6} + \frac{1}{6 \times 8} + \dots (n \text{ terms}) = \frac{k n}{n+1}$,then $k$ is equal to