Question

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

• A. $190+\omega $
• B. $192+\omega ^{2}$
• C. $190+i$
• D. $192+i$

Answer 1

Answer: (C)

$\displaystyle \sum _{k = 1}^{99} i^{k !}+\displaystyle \sum _{k = 1}^{99} \omega ^{k !}$
$\displaystyle \sum _{k = 1}^{99} i^{k !}=i^{1 !}+i^{2 !}+i^{3 !}+i^{4 !}+\ldots +i^{99 !}$
$=i-1+i^{6}+1+1+1+\ldots +1$
$=i-2+96=i+94$
$\displaystyle \sum _{k = 1}^{99} \omega ^{k !}=\omega ^{1}+\omega ^{2 !}+\omega ^{3 !}+\omega ^{4 !}+\ldots +\omega ^{99 !}$
$=\omega +\omega ^{2}+1+1+1+\ldots +1=96$
Sum $=i+94+96=i+190$