Question

If ${\sum }_{k=0}^{200}{i}^k+{\prod }_{p=1}^{50}{i}^p=x+iy$, then $(x,y)$ is

• A. $(0,1)$
• B. $(1,-i)$
• C. $(2,3)$
• D. $(4,8)$

Answer 1

Answer: (B)

$(1,-i)$

$1+i+i^2+i^3=0$

$\Rightarrow {\Sigma }_{k=0}^{200}{i}^k=i^{200}=1$

$\Rightarrow {\Pi }_{p=1}^{50}{i}^p=i^{\frac{50\left(51\right)}{2}}=i^{1275}=-i$

$\Rightarrow {\Sigma }_{k=0}^{200}{i}^k+{\Pi }_{p=1}^{50}{i}^p=1-i$

$\therefore (x,y)=(1,-i)$