$\frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \ldots$ to $24$ terms $=$

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 value of $\mathop {\lim }\limits_{n \to \infty } \left( \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2n - 1)(2n + 1)} \right)$ is equal to

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

If the sum $\frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \dots$ up to $20$ terms is equal to $\frac{k}{21}$,then $k$ is equal to

The value of the infinite product $\prod\limits_{n = 2}^\infty {\left( {1 - \frac{1}{{{n^2}}}} \right)}$ is