If $\sum_{k=1}^{10} \frac{k}{k^{4}+k^{2}+1}=\frac{m}{n}$,where $m$ and $n$ are coprime,then $m+n$ is equal to.

If the sum of the first $10$ terms of the series $\frac{1}{1+1^4 \cdot 4} + \frac{2}{1+2^4 \cdot 4} + \frac{3}{1+3^4 \cdot 4} + \frac{4}{1+4^4 \cdot 4} + \dots$ is $\frac{m}{n}$, where $\text{gcd}(m,n) = 1$, then $m+n$ is equal to:

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

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 :

For any $n \in N$,$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \ldots + \frac{1}{(3n-1)(3n+2)} = $

$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots + \frac{1}{n(n + 1)} = \dots$