If $\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots \left(1+\frac{2n+1}{n^2}\right) = 121$,then $n =$

Let $S_n = \frac{1}{1^3} + \frac{1 + 2}{1^3 + 2^3} + \frac{1 + 2 + 3}{1^3 + 2^3 + 3^3} + \dots + \frac{1 + 2 + \dots + n}{1^3 + 2^3 + \dots + n^3}$. If $100 S_n = n$,then $n$ is equal to:

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

If ${a_k} = \frac{1}{{k(k + 1)}}$ for $k = 1, 2, 3, 4, ..., n$,then ${\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)^2} = $

If $\sum_{r=1}^{25} \left( \frac{r}{r^{4}+r^{2}+1} \right) = \frac{p}{q}$ where $p$ and $q$ are positive integers such that $\gcd(p,q)=1$, then $p+q$ is equal to . . . . . . .

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 :