$\prod\limits_{n = 1}^{10} {\left( {\frac{{6\sum\limits_{i = 0}^n i + 1}}{{6\sum\limits_{j = 0}^n {(j - 1)} + 1}}} \right)} $ is equal to

If $\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m$ and $\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n$,then the point $(m, n)$ lies on the line

If $\sum_{r=1}^{n} T_{r} = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$,then $\lim_{n \rightarrow \infty} \sum_{r=1}^{n} \left(\frac{1}{T_{r}}\right)$ is equal to :

Let $S_k, k=1, 2, \ldots, 100$,denote the sum of the infinite geometric series whose first term is $\frac{k-1}{k!}$ and the common ratio is $\frac{1}{k}$. Then the value of $\frac{100^2}{100!} + \sum_{k=1}^{100} |(k^2 - 3k + 1) S_k|$ is

$1(1!) + 2(2!) + 3(3!) + \dots + n(n!) = \dots$

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 :