Question

The value of $\sum _{n=0}^{100}{i}^{n!}$ equals (where $i=\sqrt{-1})$

• A. $-1$
• B. $i$
• C. $2i+95$
• D. $97+i$

Answer 1

The correct option is C $2i+95$

$\sum _{n=0}^{100}{i}^{n!}$
$=i^{0!}+i^{1!}+i^{2!}+i^{3!}+i^{4!}+...+i^{100!}...\left(1\right)$

$i^2=-1,i^3=-i,i^4=1i^{2+4}=-1,i^{3+4}=-i,i^8=1$
The powers of $i$ which are multiple of $4$ is equal to $1$.

Therefore, from Eqn $\left(1\right)$,
$\sum _{n=0}^{100}{i}^{n!}$
$=i+i-1-1+1+1+...+1$
$=95+2i$