The value of $\sum _{\mathrm{k}=1}^{99}\left({\mathrm{i}}^{\mathrm{k}!}+{\mathrm{\omega }}^{\mathrm{k}!}\right)$ is (where, $\mathrm{i}=\sqrt{-1}$ and $\mathrm{\omega }$ is non-real cube root of unity)

• A

  $190+\omega$
• B

  $192+{\omega }^2$
• Correct answer
  C

  $190+i$
• D

  $192+\mathrm{i}$

Hence, the correct answer is option: 'C'.