In the expansion of $(1 + x + x^2)e^{-x}$,the coefficient of $x^2$ is

The value of $\sum_{r=2}^{\infty} \frac{1+2+\dots+(r-1)}{r !}$ is:

The sum of the series $\frac{1}{1 \times 2} + \frac{1 \times 3}{1 \times 2 \times 3 \times 4} + \frac{1 \times 3 \times 5}{1 \times 2 \times 3 \times 4 \times 5 \times 6} + \dots \infty$ is

Let $S_{n} = 1 \cdot (n-1) + 2 \cdot (n-2) + 3 \cdot (n-3) + \dots + (n-1) \cdot 1$,for $n \geq 4$. The sum $\sum_{n=4}^{\infty} \left( \frac{2 S_{n}}{n!} - \frac{1}{(n-2)!} \right)$ is equal to:

If $a = \sum\limits_{n = 0}^\infty {\frac{{{x^{3n}}}}{{(3n)!}}} ,\,b = \sum\limits_{n = 1}^\infty {\frac{{{x^{3n - 2}}}}{{(3n - 2)!}}} $ and $c = \sum\limits_{n = 1}^\infty {\frac{{{x^{3n - 1}}}}{{(3n - 1)!}}} $ then the value of ${a^3} + {b^3} + {c^3} - 3abc = $

$1 + \frac{2^2}{1!} + \frac{3^2}{2!} + \frac{4^2}{3!} + \dots \infty = $ (in $e$)