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

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

The sum of the infinite series $\cot ^{-1}\left(\frac{7}{4}\right)+\cot ^{-1}\left(\frac{19}{4}\right)+\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1}\left(\frac{67}{4}\right)+\ldots \ldots$ is :-

If $\frac{1}{2 \times 3 \times 4} + \frac{1}{3 \times 4 \times 5} + \frac{1}{4 \times 5 \times 6} + \dots + \frac{1}{100 \times 101 \times 102} = \frac{k}{101}$,then $34k$ 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:

$\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