If the sum of the first ten terms of the series $\frac{1}{5}+\frac{2}{65}+\frac{3}{325}+\frac{4}{1025}+\frac{5}{2501}+\ldots$ is $\frac{m}{n}$,where $m$ and $n$ are co-prime numbers,then $m + n$ is equal to

The value of $\sum_{k=1}^{13} \frac{1}{\sin \left(\frac{\pi}{4}+\frac{(k-1) \pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)}$ is equal to

The sum of the series $\frac{3}{{1! + 2! + 3!}} + \frac{4}{{2! + 3! + 4!}} + \frac{5}{{3! + 4! + 5!}} + ...... + \frac{{2008}}{{\left( {2006} \right)! + \left( {2007} \right)! + \left( {2008} \right)!}}$ is equal to

The value of $\lim _{n \rightarrow \infty} \left( \sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!} \right)$ is:

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:

Find the sum of the series $1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + \dots$ up to $n$ terms.