Question

The value of $\sum _{k=1}^{99}\left(i^{k!}+{\left(\omega \right)}^{k!}\right)$ 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)

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