• # question_answer If ${{i}^{2}}=-1$, then sum $i+{{i}^{2}}+{{i}^{3}}+...$to 1000 terms is equal to [Kerala (Engg.) 2002] A) 1 B) - 1 C) i D) 0

$i+{{i}^{2}}+{{i}^{3}}+.......$up to 1000 terms $=\frac{i(1-{{i}^{1000}})}{1-i}=$$\frac{i(1-{{({{i}^{4}})}^{250}})}{1-i}$$=\frac{i(1-1)}{1-i}=0$.