Question

If $i^2=-1$, then the value of $\sum _{n=1}^{200}{i}^n$ is

• A. 50
• B. $-50$
• C. 0
• D. 100

Answer 1

Answer: (C)

$\sum _{n=1}^{200}{i}^n=i+i^2+i^3+\dots +i^{200}$
$=\frac{i\left(1-i^{200}\right)}{1-i}$ (using sum formula of G.P.)
$=\frac{i(1-1)}{1-i}=0$