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

$\lim _{n \rightarrow \infty} \left( \frac{1}{3 \cdot 7} + \frac{1}{7 \cdot 11} + \frac{1}{11 \cdot 15} + \ldots + n \text{ terms} \right) =$

Let the greatest common divisor of $m$ and $n$ be $1$. If $\frac{1}{1 \cdot 7} + \frac{1}{7 \cdot 13} + \frac{1}{13 \cdot 19} + \dots$ up to $20$ terms $= \frac{m}{n}$,then $5m + 2n = $

The sum of the infinite series $\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \frac{4}{5!} + \dots$ is

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