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 :

Find the sum to $n$ terms of the series: $\frac{3}{1 \cdot 2} \cdot \frac{1}{2} + \frac{4}{2 \cdot 3} \cdot \left( \frac{1}{2} \right)^2 + \frac{5}{3 \cdot 4} \cdot \left( \frac{1}{2} \right)^3 + \dots$

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

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

Find the sum to $n$ terms of the series $1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \ldots$

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 $n$.