$1 + \frac{4^2}{3!} + \frac{4^4}{5!} + \dots \infty = $

  • A
    $\frac{e^4 + e^{-4}}{4}$
  • B
    $\frac{e^4 - e^{-4}}{4}$
  • C
    $\frac{e^4 + e^{-4}}{8}$
  • D
    $\frac{e^4 - e^{-4}}{8}$

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Similar Questions

$\frac{2\frac{1}{2}}{1!} + \frac{3\frac{1}{2}}{2!} + \frac{4\frac{1}{2}}{3!} + \frac{5\frac{1}{2}}{4!} + \dots \infty$ ની કિંમત શું છે?

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

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

ધારો કે $\sum_{n=0}^{\infty} \frac{n^3((2n)!) + (2n-1)(n!)}{(n!)((2n)!)} = ae + \frac{b}{e} + c$,જ્યાં $a, b, c \in \mathbb{Z}$ અને $e = \sum_{n=0}^{\infty} \frac{1}{n!}$. તો $a^2 - b + c$ ની કિંમત $................$ છે.

શ્રેણી $1 + \frac{1}{4 \cdot 2!} + \frac{1}{16 \cdot 4!} + \frac{1}{64 \cdot 6!} + \dots$ અનંત સુધીનો સરવાળો શું થાય?

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