• # question_answer If $\sum\limits_{k=0}^{100}{{{i}^{k}}}=x+iy$, then the values of $x$ and $y$are A) $x=-1,y=0$ B)  $x=1,y=1$ C) $x=1,y=0$ D) $x=0,y=1$

$\sum\limits_{k=0}^{100}{{{i}^{k}}=x+iy,}$Þ $1+i+{{i}^{2}}$$+......+{{i}^{100}}=x+iy$ Given series is G.P. Þ  $\frac{1.(1-{{i}^{101}})}{1-i}=x+iy$ Þ $\frac{1-i}{1-i}=x+iy$ Þ $1+0i=x+iy$ Equating real and imaginary parts, we get the required result.