The sum of the series $\frac{2}{2 !} + \frac{2+4}{3 !} + \frac{2+4+6}{4 !} + \ldots$ is equal to

  • A
    $e$
  • B
    $e^{-1}$
  • C
    $e^{-2}$
  • D
    $e^{-3}$

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Similar Questions

The expression $\begin{aligned} & 1+x \log _e a+\frac{x^2}{2 !}\left(\log _e a\right)^2+\frac{x^3}{3 !}\left(\log _e a\right)^3+\ldots \end{aligned}$ for $a>0, x \in R$ is equal to:

The coefficient of $x^{6}$ in the expansion of $e^{2x}$ is:

$\left( {1 + \frac{1}{{2!}} + \frac{1}{{4!}} + \dots} \right) \left( {1 + \frac{1}{{3!}} + \frac{1}{{5!}} + \dots} \right) = $

The value of the infinite series $\frac{1^{2}+2^{2}}{3 !} + \frac{1^{2}+2^{2}+3^{2}}{4 !} + \frac{1^{2}+2^{2}+3^{2}+4^{2}}{5 !} + \dots$ is:

The sum of the series $\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$ up to infinity is

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