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

For positive integers $n$,if $4 a_n = (n^2 + 5n + 6)$ and $S_n = \sum_{k=1}^n \left(\frac{1}{a_k}\right)$,then the value of $507 S_{2025}$ is :

If $\sum\limits_{n = 1}^5 {\frac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}} = \frac{k}{3}} $,then $k$ is equal to

If $\frac{1}{2 \times 7} + \frac{1}{7 \times 12} + \frac{1}{12 \times 17} + \frac{1}{17 \times 22} + \dots$ to $10$ terms $= k$,then $k =$

Find the sum of the series: $1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \dots + n \cdot n!$

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.