$\frac{1}{e^{3x}}(e^x + e^{5x}) = a_0 + a_1x + a_2x^2 + \ldots$
$\Rightarrow 2a_1 + 2^3a_3 + 2^5a_5 + \ldots$ is equal to

  • A
    $e$
  • B
    $e^{-1}$
  • C
    $1$
  • D
    $0$

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