The sum $\sum\limits_{r = 1}^{10} {({r^2} + 1) \times r!}$ is equal to

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

The sum of $(n + 1)$ terms of $\frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots$ is

The sum of the series $\frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots$ up to $(n + 1)$ terms is equal to:

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 . . . . . . .

The sum of the infinite series ${\tan ^{ - 1}}\left( {\frac{2}{{1 - {1^2} + {1^4}}}} \right) + {\tan ^{ - 1}}\left( {\frac{4}{{1 - {2^2} + {2^4}}}} \right) + {\tan ^{ - 1}}\left( {\frac{6}{{1 - {3^2} + {3^4}}}} \right) + \dots$ is