If $S_n = \frac{n(n + 1)(n + 2)}{6}$,then $\sum_{n = 1}^\infty \frac{1}{t_n} = $

The value of $\sum\limits_{k = 1}^\infty {\frac{{3{k^2} + 3k + 1}}{{{{\left( {{k^2} + k} \right)}^3}}}} $ is equal to

$\sum_{k=0}^{12} \frac{1}{\sin \left((k+1) \frac{\pi}{6}+\frac{\pi}{4}\right) \sin \left(\frac{k \pi}{6}+\frac{\pi}{4}\right)} = $

Let for $n = 1, 2, \ldots, 50$,$S_{n}$ be the sum of the infinite geometric progression whose first term is $n^{2}$ and whose common ratio is $\frac{1}{(n+1)^{2}}$. Then the value of $\frac{1}{26} + \sum_{n=1}^{50} \left(S_{n} + \frac{2}{n+1} - n - 1\right)$ is equal to

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

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