The sum of the series $1 + \frac{3}{2!} + \frac{7}{3!} + \frac{15}{4!} + \dots \text{to } \infty$ is

  • A
    $e(e + 1)$
  • B
    $e(1 - e)$
  • C
    $3e - 1$
  • D
    $e(e - 1)$

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$1.5 + \frac{2.6}{1!} + \frac{3.7}{2!} + \frac{4.8}{3!} + \dots$ is equal to

If $x=1+\frac{1}{2 \times 1 !}+\frac{1}{4 \times 2 !}+\frac{1}{8 \times 3 !}+\ldots$ and $y=1+\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}+\frac{x^{6}}{3 !}+\ldots$, then the value of $\log_{e} y$ is

$1 + x \log_e a + \frac{x^2}{2!} (\log_e a)^2 + \frac{x^3}{3!} (\log_e a)^3 + \dots = $

$\frac{1}{1!} + \frac{1 + 2}{2!} + \frac{1 + 2 + 2^2}{3!} + .....\infty = $

$\frac{e^2 + 1}{2e} = $

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