Q. $i^{2} + i^{4} + i^{6} + ...upto\ (2k + 1)terms,\ k\epsilon N$ is
A) 0
B) 1
C) $- 1$
D) $k$
Explanation
(c)
$i^{2} + i^{4} + i^{6} + \ldots$upto $(2k + 1)$ terms
$= \frac{i^{2}\lbrack 1 - \left( i^{2} \right)^{2k + 1}\rbrack}{1 - i^{2}} = \frac{- 1\lbrack 1 - ( - 1)^{2k + 1}\rbrack}{1 + 1} = \frac{- \lbrack 1 - ( - 1)\rbrack}{2} = - 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.